Question

A 6.30kg block is pushed 9.45m up a smooth 37.0? inclined plane by a horizontal force of 80.0N . If the initial speed of the block is 3.35m/s up the plane.

Part A

Calculate the initial kinetic energy of the block.

Part B

Calculate the work done by the 80.0N force.

Part C

Calculate the work done by gravity.

Part D

Calculate the work done by the normal force.

Part E

Calculate the final kinetic energy of the block.

Answer 1

initial kinetic energy = 1/2 m v1^2 = 1/2 x 6.3 x (3.35)^2 = 35.351 J

work done by 80N force >> (force) x (displacement in direction of force) = (80) x (9.45 cos37 ) = 603.768 J

work done by gravity >> (gravity force) x (displacement in direction of force) = (6.3 x 9.8) x (9.45 x sin 37) = 351.124 J

work done by normal force >> it will be 0 as displacement is always perpendicular to dorection of force

final kinetic energy >> by work energy theorem >> change in kinetic energy = work done by all forces

KE (final) - 35.351 = 603.768-351.124 >> KE(final) = 287.995 J

Answer 2

initial Ke = .05mv^2 = 1/2*6.3*3.35^2 = 35.35

net force on block = Fcos(37) - mgsin(37)

net a = (80*cos(37) - 6.3*9.8*sin(37))/6.3 = 4.24m/s^2

work done = fcos(37)*9.45 = 80*cos(37)*9.45 = 603.92 J

work done by gravity = -mgsin(37)*9.45 = -6.3*9.8*sin(37)*9.45 = -350.97 J

work done by normal force = 0

final v

v^2 = 3.35^2 + 2*4.24*9.45

1/2mv^2 = .5*6.3*(3.35^2 + 2*4.24*9.45) = 287.78J