Box is sliding down a slide. Speed, coefficients of friction help?

Question

A 20 kg box is released from the top of a long, frictionless slide 12 m above the ground. Upon reaching ground level, it slides for a distance of 7.0 m before reaching point Q. Find the speed of the box when it reaches a height of 6 m above the ground. Find the speed of the box when it first reaches ground level. Find the value for the coefficient of friction that would cause the box to come to rest at point Q.

Thanks for any help! =)

Answer 1

<< Find the speed of the box when it reaches a height of 6 m above the ground.>>

Applying the law of conservation of energy,

Potential energy @ 12 m = Potential energy @ 6 m + Kinetic energy

Potential energy = PE = mgh

Kinetic energy = (1/2)mV^2

where

m = mass of the box

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

h = vertical distance of the box's position

V = velocity of the box

Substituting values,

(20)(9.8)(12) = (20)(9.8)(6) + (1/2)(20)(V^2)

Rearranging,

V^2 = 2*9.8(12 - 6)

V^2 = 117.6

V = 10.84  m/sec --- this is the speed of the box when it reaches a level of 6 meters above the ground

<< Find the speed of the box when it first reaches ground level.>>

For this, the law of conservation of energy still applies.

Potential energy at 12 m high = Kinetic energy when box hits bottom

m(9.8)(12) = (1/2)(m)V^2

V^2 = 2*9.8*12 = 235.2

V = 15.34 m/sec.

<< Find the value for the coefficient of friction that would cause the box to come to rest at point Q.>>



To determine the coefficient of friction, the acceleration of the box must first be determined.

Use the formula

Vf^2 - V^2 = 2as

where

Vf = velocity of box at point Q = 0 (since it will stop)

V = 15.34 (as calculated above)

a = acceleration

s = stopping distance = 7 m (given)

Substituting values,

0 - 15.34^2 = 2(a)(7)

a = - 16.81 m/sec^2

The negative sign attached to the acceleration simply indicates that the box is slowing down as it heads toward point Q. HOWEVER,  I will stop my analysis here. The answer for acceleration that I have is a ridiculously high number. This just cannot be so. There must be an error in some of the data given.

I suspect that the either the stopping distance of 7 or the initial height of 12 meters is erroneous.

Answer 2

Well that depends on if the box is wet or not.