Question

an object of mass 2kg is released from a top of inclined plane 30° and height 6m. The coefficient of kinetic friction of the surface is 0.15. a , What is the work done against fraction when the reaches the bottom ? b , What is the kinetic energy of the object when it reaches the bottom ? c , What is the speed of the object when it reaches the the bottom ?​

Answer 1

Explanation:

1) Work done = force x distance x cos(θ)

= 0.15 x 6 x cos(30)

= 0.779

2) Ek = ½mv²

v = acceleration due to gravity so 9.81

Ek = ½(2)(9.81)²

Ek = 96.2361

3) v = (√(2em)) / m

= (√(2(96.2361)(2)) / 2

= 9.81 so especially with no time given, I can only assume the acceleration due to gravity but take it with a pinch of salt.