Physics Question: Dynamics?
A 45 kg woman is standing in an elevator that is accelerating downwards at 2.0 m/s2. What force (normal force) does the elevator floor exert on the woman's feet during this acceleration? A. 90 N B. 350 N C. 440 N D. 530 N
Physics dynamics question!?
according to a simplified view of Newton´s second law F = M A in this case, the ecuation holds as long as the mass remains constant. M = F/A but the acceleration given has no units! asuming they are m/s^2 (this is unsolvable without a unit assumption) then the answer to the question becomes M = .35 / .15 which gives us a mass of 2 1/3 Kg
Physics Dynamics Question?
Downslope component of gravity = mg sin (theta) Normal force = mg cos (theta) Friction (pointing upslope) = mu mg cos(theta) So acceleration is the net force over the mass: a = g (sin (theta) - mu cos(theta)) Since you started from rest, your final velocity is: vf = at = gt (sin(theta) - mu cos(theta) )
Physics Dynamics Question?
I will assume by impending motion you mean that if the truck had accelerated at any rate faster than it did the load would have slid across the deck of the trailer. a. Determine the friction coefficient between the crate and the bed. Let C equal the mass of the crate in kilograms. The vertical force down would therefore be g*C=9.8*C newtons The horizontal force would be the acceleration of the truck time the mass of the crate. meters per hour/time of acceleration in seconds/seconds per hour 100000/12/3600*C newtons=2*C The ratio of the horizontal and vertical forces it the coefficient of friction 2*C/(9.8*C)=2/9.8=0.204 notice that coefficient has no units. b. Determine the maximum acceleration up an 8 degree incline so that the crate does not slide off the truck. Traveling on an incline reduces the force perpendicular to the bed of the truck. The force due to gravity has a component Parallel to the bed of the truck and therefore increases the force parallel to the bed of the truck experienced by the crate at any particular acceleration of the truck. The coefficient of friction does not change. The force perpendicular to the bed of the truck is proportional to the cosine of the angle of incline. cos(8degrees) times 9.8*C=Fperp. Fperp*coef=maximum force of resistance ot motion 9.8*C*.99*.204=1.98*C newtons The force Parallel to the bed of the truck due to gravity is proportional to the sine of the angle of incline Sin(8 degrees)times 9.8*C=parallel gravitational force .139*9.8*C=1.36*C That leaves 1.98*C-1.36*C=.62*C of resistance or an acceleration of .62 Meters per Second per Second or an acceleration of 2.2 Kilometers per Hour per Second
How to solve physics question (Dynamics)?
A 12-kg turtle rests on the bed of a zookeeper’s truck, which is traveling down a country road at 55 mph. The zookeeper spots a deer in the road, and slows to a stop in 12 s. Assuming constant acceleration, what is the minimum coefficient of static friction between the turtle and the truck bed surface that is needed to prevent the turtle from sliding
How do I solve this physics question (dynamics)?
Q: How do I solve this physics question (dynamics)? https://imgur.com/Vg2Z4siA: You will recall the force needed to accelerate a mass is f = m*am to be accelerated is (5 + x) kgA drawing will show there is kinetic friction force acting downwardsf = 0.3 * 80 newtons = 24 newtonsThere is also the 5 kg weight pulling downwards with 5*g = 5*9.81 = 49.05NThe accelerating force upwards is provided by the force of weight x * 9.81 N LESS the kinetic friction force 24N, LESS the weight of the 5 kg mass = 5 * 9.81 NThis balance of forces is 8+ 1.6*x = 9.81*x - 24 - 49.05which boils down to x = 81.05 / 8.21 = 9.872 kg.Looking at the multiple choices a) 9.9 kg appears optimal!
Impulse question [Dynamics] (physics)?
car's mass = 15,680/9.81 = 1598.4 kg = m initial speed = 20.0 m/s final speed = 0 change in speed = -20.0 m/s {neg sign indicates a reduction in speed} momentum change= -20.0m = -20.0(1598.4) = -31,968 kg-m/s ≈ -32,000 kg-m/s ANS
Can someone help with this physics question about dynamics?
A 66.2 kg spacewalking astronaut pushes off a 646.0 kg satellite, exerting a 187.0 N force for the 0.558 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 3.77 min?
Challenging Physics Question on Dynamics! Help?
A painter is in a crate, which hangs alongside a building. There is a rope which is connected vertically to the crate on the outside, around a pulley and back to the inside of the crate. The painter can pull this rope to influence the acceleration of the crate( and him). Take g=10m/s^2. When the poainter who weighs 1000 N, pulls on the rope, the force he exerts on the floor of the crate is 450N. The crate weighs 250N. Find the acceleration of the system?