Abox 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

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