A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of ki
The force of friction = coefficient * normal force. Normal force = mass * gravitational acceleration. Substituting, Ffr = coefficient * mass * gravity. We know the coefficient and gravity, so plugging in: Ffr = 0.21*-9.8*m. That's the sum of forces, which is equal to mass * acceleration. Guess what? The masses cancel out. 0.21*-9.8*m = ma a = 0.21*-9.8 = -2.058 m/s^2 (acceleration is negative; the box is slowing down) Using v^2 = (v0)^2 + 2*a*x to solve for x. v^2, desired velocity, is going to be zero. You're solving for when it stops. v0^2 is 4.0 m/s, given. a = -2.058 m/s^2, you just calculated that. -16 = -4.116x x = 3.887 meters
What are a house cat's favorite toys?
Ping pong balls, bounce them and they leap to catch them (miss every time) and then bat them around the floor. Of course, you will spend time retrieving them from under furniture, refrigerator…When we fostered some kittens, we put the kittens in the bath tub and added a ping pong ball and had kitty soccer.Paper bags, boxes, hair ties (they steal them from my makeup bag), small stuffed animals (they carry them around in their mouth, hug and kick them with back paws) add cat nip spray to stuffed animals.And the must have scratching post, multiple levels and a hidey hole.
Why does my cat love balled up paper?
Balled up paper makes a crinkly sound, which many cats love. Just be happy that your kitty is easy to please ;P
Potential energy and energy conservation- loop-the-loop?
we will use newton's second law and energy conservation the car will rise to a height h above the floor; the car travels on the inside of the track call theta the angle between the a radius of the circle and the vertical when the car is moving up on the inside of the track, we write newton's second law for forces acting radially inward the normal force points radially inward the component of gravity along a radius is mg cos(theta) these forces combine to produce an inward centrip accel, these give us -N - mgcos(theta) = -mv^2/R or N=mv^2/R - mgcos(theta) the car leaves the track when N=0, so the condition for leaving the track is v^2 = g R cos(theta) eq. 1 now we use conservation of energy to find an expression for v^2 as a function of R and theta at the top, all the energy is in PE, and this amount of PE is 2mgR when the car is at a height h above the bottom, it has both KE and PE; the KE is 1/2mv^2 and the KE is mgh we now find an expression for h assuming the car will leave the track when its height is greater than R, we can write h as h=R+Rcos(theta)=R(1+cos(theta)) so we have 1/2 mv^2+mgR(1+cos(theta))=2mgR 1/2 v^2 = 2gR-gR(1+cos(theta)) v^2=2gR(1-cos(theta)) substitute this value of v^2 into eq. 1: 2gR(1-cos(theta)=gR cos(theta) cos(theta)=2/3 therefore, h=R(1+cos(theta))=5/3 R and this is the height when the car leaves the track
What is the weirdest thing your boss has asked you to do?
My first job at a retail clothing store in a mall. My boss/manager asked me to go to the toy store and pick up something for his nephew…on company time.I was young and didn’t question the appropriate-ness: was this my job? Should I be doing this on company time?…Months later, the manager was fired for embezzling 20–30K US$ from the company. The cash he gave me on the day of the toy store purchase was most likely stolen money.That was not only weird, it was wrong. I don’t feel terribly bad about it, because I was young and inexperienced. I told the company higher-ups what had happened and they were fine with my actions.Take a deceased patient to the hospital morgue by myself at night. I worked at a hospital as a “messenger”, which was really a title for “do everything we ask you to do, do it now, and do it quickly.” It was about 10pm at night, and a patient had died in the ICU. I received a call, and given the order to take patient X to the morgue in the basement. “OK”, I said, “no problem.” But, it was a problem. I was 18 years old and weighed about 170 pounds. The deceased patient weight about 300 pounds. Let me back-up and explain that many hospitals do not have the gigantic morgue rooms that you may have seen on TV.Typically, it is a small room with sliding drawers from floor to ceiling to temporarily keep the deceased cold until the funeral home comes to pick them up. The challenge is to get the body on the sliding tray. One can adjust the height of a gurney, but that is only half the job. One needs to slide them over. That’s a huge challenge, when you are alone.I went back to the nurses on the ICU floor and asked for help. They said they were short-staffed and couldn’t help at the moment; please find someone else to help. I went to the only person I knew who could help: a friend from high school who worked in the kitchen. He hasn’t been the same since that trip to the morgue at night and still brings it up as a terrible and haunting experience!
Why won’t my cat play with toys?
Sometimes it’s the simple things…A rolled up sheet of paper, a ping pong ball, a plastic bottle top - or the plastic ring that is sometimes below the bottle cap, the cardboard roll that’s left from a roll of toilet paper… anything that can roll (or slide on a hard floor) could end up being a favourite toy. Some cats will readily play with toys that you buy, others will ignore those expensive toys & just jump into the box that said toys came in. Most cats can’t resist a cardboard box.If your cat is allowed outside, a gentle breeze can make fallen leaves a toy for a while.
If a block of mass (m) is kept on a horizontal surface with friction coefficient [math]\mu[/math], then what would be the minimum force needed to move that block on the surface?
This is actually a tricky one since many people will end up gettingminimum force as [math]f = \mu.mg [/math]However Just read the question again , it says minimum force (There is no mention of minimum horizontal force)Now as we know Friction [math]force \sim Normal[/math]i.e. : if we minimize the normal reaction we can minimize the friction at the same time we also need to provide enough Horizontal force to ensure block gets a horizontal velocitySomething like shown in figure !Now the task is to find optimal angle “x” and Magnitude FApplying simple physics : (Assuming Normal reaction from surface is N)Balancing Vertical force[math]Fsinx+N = mg [/math] —>(1)[math]N = mg - Fsinx[/math]Balancing Horizontal force[math]Fcosx = Friction[/math]=> [math]Fcosx = \mu.N[/math]=> [math]Fcosx = \mu.(mg-Fsinx) [/math]=> [math]F(cosx+\mu.sinx) = \mu.mg [/math]=> [math]F = \frac{\mu.mg}{cosx+\mu.sinx} [/math] ==>(2)Now to minimize F(2) we need to maximize denominator of 2[math]max[/math]{[math]cosx+\mu.sinx[/math]}[math] = \sqrt{1+(\mu)^2} [/math] Basic TrigoSo minimum force needed would be[math]F = \frac{\mu.mg}{\sqrt{1+(\mu)^2}}[/math]
Physics College Question. Block 1, with mass m1 and speed 4.8 m/s, slides along an x axis on a frictionless floor and then undergoes a....?
Conserve momentum: initial p = final p m1*4.8m/s + 0 = m1*u + 0.3m1*v for u, v the post-collision velocity of m1, m2 respectively. m1 cancels, leaving 4.8 = u + 0.3v For an elastic, head-on collision, we know (from CoE) that the relative velocity of approach = relative velocity of separation, or 4.8 m/s = v - u v = u + 4.8 → plug that into the momentum equation 4.8 = u + 0.3(u + 4.8) = 1.3u + 1.44 u = (4.8 - 1.44)/1.3 = 2.6 m/s v = 2.6 + 4.8 = 7.4 m/s KE1 = ½*m1*2.6² = m1*3.34m²/s² KE2 = ½*m2*7.4² = m2*27.3m²/s² work1 = KE1 = µmgd = 0.5*m1*9.8m/s²*d1 = m1*3.34m²/s² → d1 = 0.68 m ◄ (A) work2 = KE2 = 0.5*m2*9.8m/s²*d2 = m2*27.3m²/s² → d2 = 5.6 m ◄ (B) If you find this helpful, please select Best Answer!