A 12.0kg box initially at rest slides from the top of a ramp.The top of the ramp is 6.0 m above the ground.If the force of friction along the 11.0 m long incline is 5.0 N.

Question

A 12.0kg box initially at rest slides from the top of a ramp.The top of the ramp is 6.0 m above the ground.If the force of friction along the 11.0 m long incline is 5.0 N, what is the speed of the box as it reaches the bottom of the ramp?

Answer 1

The component of gravity acting down the ramp is mg(sinϴ). For this problem, that is

(12.0 kg)(9.81 m/s²)(6/11) ≈ 64.2 N

Subtracting friction, the net force is 59.2 N

The net work done on the box is F_net d = (59.2 N)(11.0 m) = 651 joules

This work results in kinetic energy: 651 J = (1/2)mv² = (6.0 kg)v²

^'Solving this gives v ≈ 10.4 m/s

Answer 2

Jo

The net force on the box down the ramp is mg sin Θ - 5 (mg sin Θ is the force down the ramp due to gravity, 5 is the force due to friction up the ramp, m is the mass of the box, g= 9.8 m/s/s, and Θ is the angle of inclination of the ramp above the horizontal). m = 12 kg, sin θ = 6/11 (because the top of the ramp is 6 m above the ground, and the incline is 11 m long).

So using F=ma, F= mg sin Θ - 5, we have ma = mg sin Θ - 5. 12 a = 12 . 9.8 . (6/11) - 5 giving a = 4.93 m/s/s down the ramp.

Using v² = u² + 2as (with u=0 the initial speed, a= 4.93 , and s =11) gives v = 10.4 m/s

This will be the speed of the box at the bottom of the ramp.

(To be completely accurate you should use an exact value for a in this last calculation substituting an expression for a [ 9.8 . (6/11) - (5/12) into your calculator . ]

Mike