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

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