Question

A 5.0 kg box slides down a 5.0 m high frictionless hill, starting from rest, across a 2.0 m wide horizontal surface, then hits a horizontal spring with spring constant 500 N/m. The ground under the spring is frictionless, but the 2.0 m wide horizontal surface is rough with a coefficient of kinetic friction of 0.25.

a. What is the speed of the box just before reaching the rough surface?

b. What is the speed of the box just before hitting the spring?

c. How far is the spring compressed?

Answer 1

let's startbwith calculaing teh velocity of mass at the bottom of slide using conservation of energy,

mgh = 1/2 mv^2

v= sqroot ( 2gh) = sqroot ( 2 x 9.8x 5) = 9.899 m/s apprx

a) speed  speed of the box just before reaching the rough surface= 9.899 m/s apprx

b) retdartion expereineced by block owing to rough surface= 0.25. (9.8) = 2,45 m/s^2

speed of the box just before hitting the spring = v^2 = 9.899^2 - 2 (  2,45) ( 2) =97.99 -9.8

v = 9.39 m/s apprx

c) using the conservation of energy

1/2mv^2 = 1/2 kx^2

5( 88.19) = 500 (x)^2

x( spring compressed) =0.939 m apprx