Question

A block of mass 4.0Kg slides down an incline plane of length 10 meters that makes an angle of 30 degrees with the horizontal. The coefficient of kinetic friction between the block and the incline is 0.3. If the block is has an initial speed of 2mis down the incline at the top of the incline, then what is the speed at the bottom? Show calculations. Indicate answer.

In the previous problem, what was the gain in Kinetic Energy? Show calculation. Indicate answer. What was the change in potential energy? What was the work done by friction?

Answer 1

Friction = f = μ_kN , where N is the normal reaction

=> f = (μ_k)(m*g*cosθ) = (0.3)(4*9.81*cos30) = (0.3)(4*9.81*0.867) = 10.21 N

Let 'a' be the acceleration of the block

(m*g*sinθ) - f = ma

=> (4*9.81*sin30) - 10.21 = 4a

=> (4*9.81*0.5) - 10.21 = 4a

=> a = 2.35 m/s²

s = ut + (1/2)at²

=> 10 = (2)(t) + (0.5)(2.35)t²

=> 1.175t² + 2t -10 = 0

=> t = 2.19

v = u + at

=> v = 2 + (2.35)(2.19)

=> v = 7.15 m/s

Gain in kinetic energy = (1/2)mv² - (1/2)mu²

=> ΔKE = (1/2)*m*(v² - u²)

=> ΔKE = (1/2)*4*(7.15² - 2²)

=> ΔKE = 94.24 J

Work done by friction (W) = -f*s (negative sign indicates that the force is actiong opposite to displacement)

=> W = -(10.21)(10)

=> W = -102.1 J

W = ΔKE + ΔPE

=> -102.1 = 94.24 + ΔPE

=> ΔPE = -196.34

negative sign indicates that the potential energy decreases as the moves down the inclined slope