Question

A 6.20 kg block is pushed 9.50 m up a smooth 36.0 ∘inclined plane by a horizontal force of 70.0 N . If the initial speed of the block is 3.50 m/s up the plane.

Calculate the initial kinetic energy of the block.

Calculate the work done by the 70.0 N force.

Calculate the work done by gravity.

Calculate the work done by the normal force.

Calculate the final kinetic energy of the block.

Please explain

Answer 1

a)

Initial Kinetic energy, KE = 0.5*mv^2

= 0.5*(6.2)*(3.5)^2

= 38 J

b)

Work done by 70 N force = F*cos(36 deg)*d

= 70*cos(36 deg)*9.5

= 538 J

c)

Work done by gravity = -mg*sin(36 deg)*d

= -6.2*9.8*sin(36 deg)*9.5

= -339.3 J

d)

Work done by normal force = 0 <---- as the displacement is perpendicular to the normal force

e)

Final KE of the block = initial KE + Net work done on the body

= initial KE + Work done by 70 N force + Work done by gravity + Work done by normal force

= 38 + 538 - 339.3 + 0

= 236.7 J