Physics: A 8.50-g bullet embeds itself in a 0.535-kg block, which is attached to a spring of force constant...?
...43.20 N/m. If the maximum compression of the spring is 1.38 cm, find the following. (a) the initial speed of the bullet __m/s Could you help me explain these clues and how they were found: a) initial speed = (max displacement * square root [spring constant * (bullet mass + block mass)])/bullet mass b) time for system to come to rest = period / 4 given the below equations: Equations that can be possibly used: PERIODIC MOTION: Frequency: f = 1/T Angular frequency: ω = 2πf = 2π/T SIMPLE HARMONIC MOTION: Position versus time: x = Acos[(2π/T)(t)] = Acos(ωt) COONECTIONS BETWEEN UNIFORM CIRCULAR MOTION AND SIMPLE HARMONIC MOTION: Velocity: v = - Aωsin(ωt) Acceleration: a = -Aω²cos(ωt) Maximum speed: v_max = Aω Maximum acceleration: a_max = Aω² THE PERIOD OF A MASS ON A SPRING: Period of a mass and a spring: T = 2π(√m/k) EDIT: Well, the bullet block system starts at rest = 0m (equilibrium position) and works its way to the outward most displacement (of the spring) = 0.0138 meters. So I guess I understand part b) and the time for the system to come to rest. I really don't understand how my fellow student came up with the equation he did for part a)
S= ut + ½at²here the sandbag got the initial velocity of 3m/s upwards.=( -5 ×3 ) + ½ ×10× 5² m= −15 + 125 m= 110mHope it’s clear
The premise of your question is false. Newton's First Law says only that in the absence of forces, the velocity of an object is constant. But as Robert Frost already pointed out, a rocket traveling in space does encounter forces: gravitational forces from the sun and other planets.Besides: with the huge distances we are talking about, even if we did not have to worry about gravitational forces, to get anywhere within our lifetimes, we need to accelerate the rocket to quite a high speed in a short time. That takes a lot of fuel.
An object moving in acircle of radius rwithconstant speed vis accelerating. The direction of its velocity vector is changing all the time, but the magnitude of the velocity vector stays constant. The acceleration vector cannot have a component in the direction of the velocity vector, since such a component would cause a change in speed. The acceleration vector must therefore be perpendicular to the velocity vector at any point on the circle. But what is its magnitude?Let us compute the average acceleration of the object as it moves from position 1 to position 2, and its velocity changes from v1tov2. The travel time ∆t is the distance traveled divided by the speed. The distance traveled is 2θr, with the angle θ measured in radians, therefore ∆t = 2θr/v. The change in velocity ∆v is 2vsinθ pointing towards the center of the circle. The average acceleration therefore is a= ∆v/∆t = v2sinθ/(rθ) towards the center of the circle. For small angles θ, sinθ ~ θ, if θ is measured in radians.The instantaneous acceleration a is therefore given by, whereis a unit vector pointing towards the center of the circle. This acceleration is calledradialacceleration orcentripetal acceleration.
Physics: A block is hung by a string from the inside roof of a van. When the van goes straight ahead at a....?
Let's work in the frame of reference of the van (moving at a constant speed along an unbanked curve). The forces acting on the block are a. Its weight mg vertically downward b. Pseudo force (mv^2)/r radially outward c. Tension T along the string. The angle the string makes with the vertical is @. Resolve the forces in vertical and horizontal direction and apply newton's first law. You'll get Tcos@=mg and Tsin@=(mv^2)/r. Divide second equation by first. You'll get tan@=v^2/rg. :-)
Frictional force (f) will try to reduce the speed of the block. And hence the block will eventually stop. We can apply the law of conservation of energy to find out the distance travelled by the block before it stops.The energy used by frictional force decreases the kinetic energy (KE) of the block.So, change in KE of the block = Work done by friction.Change in KE = 0.5 * m * (v^2 - u^2) = (-50 * m)where, v = final speed = 0 m/su = initial speed = 10 m/sm = mass of blockWork done by friction = f * s * cos(180) = (0.1 * m * g) * s * (-1) = (-m * s)s = distance travelled before the block stopped.work is calculated by the dot product of force and displacement. Since the force is acting in backward direction and the displacement is in the forward direction, the angle between them is 180.Equating both,-m * s = -50 * m=> s = 50 meters
the mass of block is 10 kg .so it produce 100 newton normal force on earth .the coefficient of kinetic friction is 0.5 (here friction is kinetic).so the frictional force is 50 N. the external force is 5t. so net force on particle is =5t-50.we know net force = mass*acceleration5t-50=10*aso a= 0.5t-5we know acceleration is dv/dt so0.5t-5=dv/dtdv=(0.5t-5)dtdv=0.5t dt-5 dtintegrate both side, (i cant show integration sign i am using laptop)v= o.5 t2/2–5t ( t2 is t square)put time=20 sec in above linev=0.5(400)/2–0.5(20)v=90m/sec
Help solve this ferris wheel problem?
a) Angular velocity w = 3 rev/min = 3 * 2pi/60 rad/s = pi/10 rad/s Or w = 0.314 rad/s Radius r = 19.0 m/2 = 9.5 m Centripetal acceleration a = w^2 * r = 0.314^2 * 9.5 = 0.937 m/s^2 b) Mass M = 40.0 kg Let force = F F is upward and weight Mg is downward. Centripetal force is upward. F - Mg = Ma Or F = M(a+g) = 40 * (9.8 + 0.937) = 429.5 N c) Mass M = 40.0 kg Let force = F F is upward and Mg is downward. Centripetal force is downward. Mg - F = Ma Or F = M(g - a) = 40 * (9.8 - 0.937) = 354 N d) Let force = F at angle theta below vertical. Vertical component of F = F cos(theta) upward. Horizontal component of F = F sin(theta) There is no acceleration in vertical direction. Therefore F cos(theta) = Mg--------------------------(1) Centripetal force is horizontal. Therefore F sin(theta) = Ma---------------------------(2) Dividing (2) by (1), tan(theta) = a/g Or theta = atan(a/g) = atan(0.937/9.8) = 5.46 deg From (1), F = Mg/cos(theta) = 40 * 9.8/cos(5.46 deg) = 394 N Therefore, force by the seat is 394 N at 5.46 deg below vertical. (Note: I do not know what exactly is meant by inward from vertical. I have calculated the angle from vertically upward. If you want the angle from vertically downward, then it is 180 - 5.46 = 174.54 deg
s = u*t + (1/2)*a*(t^2)s = 25 m, u = -7 m/s, a = 9.8 m/s^2 Solve for t. I hope you know how to solve quadratic equations.All the best with your homework