Answer:
x=2.21m
Step-by-step 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