A box of mass 15.1 kg with an initial velocity of 2.33 m/s slides down a plane, inclined at 23◦ with respect to the horizontal. The coefficient of kinetic friction is 0.56. The box stops after sliding a distance x. How far does the box slide? The acceleration due to gravity is 9.8 m/s 2. The positive x-direction is down the plane. Answer in units of m

x=2.21m

Explanation:

A box of mass 15.1 kg with an initial velocity of 2.33 m/s slides down a plane, inclined at 23◦ with respect to the horizontal. The coefficient of kinetic friction is 0.56. The box stops after sliding a distance x. How far does the box slide? The acceleration due to gravity is 9.8 m/s 2. The positive x-direction is down the plane. Answer in units of m

Energy lost by box = work done by friction.

If the box stops after distance x, it has lost potential energy of mgx. sinθ and Kinetic energy of (1/2) mVo^2.

F=μN

coefficient of kinetic friction is the ratio of frictional force to the normal reaction

The normal reaction from the plane is mgcosθ,

friction force = μmg. cosθ, and

since work done is the product of force and distance covered

the work done by friction is μmgx. cosθ.

So mgx. sinθ + (1/2) mVo^2 = μmgx. cosθ

(1/2) mVo^2 = mgx (μ. cosθ - sinθ)

x = Vo^2/2g (μ. cosθ - sinθ)

slotting the parameters into the equation above, we have

m=15.1kg

velocity=2.33m/s

angle θ=23deg

The coefficient of kinetic friction is 0.56.

the acceleration due to gravity g = 9.8 m/s^2

x=2.33^2 / (2*9.81 (0.56. cosθ23 - sin23))

x=2.21m