Question

A 2-kg block is dragged over a rough horizontal surface by a constant force of 15 N acting at an angle of 60° above the horizontal. The speed of the block increases from 4.0 m/s to 8 m/s in a displacement of 9 m. What work was done by the friction force during this displacement?

Need quick help please

Answer 1

Net horizontal force = mass *acceleration = 2 a
goes 9 meters in time t
v = Vi + a t = 4 + at = 8
so
a t = 4 and a = 4/t
distance= 4 t + (1/2) a t^2 = 9
4 t + (1/2)(4/t) t^2 = 9
4 t + 2 t = 9
t = 1.5 seconds
a = 4/1.5 = 2.67 m/s^2
===================
net horizontal force = m a = 2 * 2.67 = 5.34 Newtons
work done by net horizontal force = 5.34 * 9 = 48.06 Joules
work done by the puller = 15 cos 60 * 9 = 67.5 Joules
so
difference is work turned to heat by friction 67.5 - 48.1

Answer 2

A 2-kg block is dragged over a rough horizontal surface by a constant force of 12 N acting at an angle of 60° above the horizontal. The speed of the block increases from 4.0 m/s to 11 m/s in a displacement of 9 m. What work was done by the friction force during this displacement?