A block is released from rest at the top of a frictionless inclined plane 16 m l

Question

A block is released from rest at the top of a frictionless inclined plane 16 m long. It reaches the bottom 4.2 s later. A second block is projected up the plane from the bottom at the instant the first block is released in such a way that it returns to the bottom simultaneously with the first block. a) find the acceleration of each block on the incline. b)what is the initial velocity of the second block? c) how far up the incline does it travel? you can assume that both blocks experience the same acceleration

Explanation / Answer

How to find acceleration, initial velocity and distance? A block is released from rest at the top of a frictionless incline plane 16m long. It reaches the bottom 4.2s later. A second block is projected up the plane from the bottom at the instant the first block is released in such a way that it returns to the bottom simultaneously with the first block. a. find the acceleration of each block on the incline. b. what is the initial velocity of the second block? c. how far up the incline does it travel? it can assumed that both blocks experience the same acceleration. for the first two questions we will use the uniform acceleration formula: X = X0 + V0*t + (a*t^2) / 2 let's define the top of the plain as X0 = 0. the far end of the slope will be X = 16 m. the first block is released from rest so V0 = 0 too. when we put in the equation the values X = 16m, t= 4.2 s we can solve for a: a = (2*X) / t^2 = 32 m/ 17.64 sec^2 = 1.81 m / sec^2 to find the initial speed of block no. 2 we'll use the fact that at time t = 4.2 sec it returns to it's initial position X = X0 so they cancel each other out of the equation. this time let's solve the equation for V0: V0 = (a*t) / 2 = (1.81 * 4.2) / 2 = 3.81 m/sec (this time either V0 or a needs to be negative in the equation because the initial speed of the second block is up the plain while it's acceleration is down the plain) for the third and final question we will use another formula: V^2 = V0^2 - 2*a*(X-X0) where V is the speed at position X. in our case the final speed is 0 (the body comes to rest at it's highest point and then start rolling back down). Vo and a are known and X0 = 0 so we just need to solve for X: X = v0^2 / (2*a) = 4.01 m

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