Part A How much work is done by the horizontal force F P = 150 N on the 18- k g
ID: 2259483 • Letter: P
Question
Part A How much work is done by the horizontal force FP = 150 N on the 18-kg block of the figure(Figure 1) when the force pushes the block 4.4m up along the 32? incline? Assuming a coefficient of friction ?k = 0.10. [Hint: Work-energy involves net work done.] Express your answer to two significant figures and include the appropriate units.Part B How much work is done by the gravitational force on the block during this displacement? Express your answer to two significant figures and include the appropriate units.
Part D What is the speed of the block (assume that it is zero initially) after this displacement? Express your answer to two significant figures and include the appropriate units.
Part A How much work is done by the horizontal force FP = 150 N on the 18-kg block of the figure(Figure 1) when the force pushes the block 4.4m up along the 32? incline? Assuming a coefficient of friction ?k = 0.10. [Hint: Work-energy involves net work done.] Express your answer to two significant figures and include the appropriate units.
Part B How much work is done by the gravitational force on the block during this displacement? Express your answer to two significant figures and include the appropriate units.
Part D What is the speed of the block (assume that it is zero initially) after this displacement? Express your answer to two significant figures and include the appropriate units.
Part B How much work is done by the gravitational force on the block during this displacement? Express your answer to two significant figures and include the appropriate units. Part D What is the speed of the block (assume that it is zero initially) after this displacement? Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
Work done by force FP = Force x displacement in the direction of force
= 150 x 4.4 = 660 Joules
Work done by gravity = Force x displacement in vertical direction
= 18 x 9.81 x (-4.4 sin32)
= - 411.72 Joules....(- sign signifies displacemnet in opposite direction)
Net force on the body = FP - mgsin32 - u(mgcos32)
= 150 - 18*9.81(0.6147)
= 41.452 Newton
acceleration, a = F/m = 2.30 m/s^2
final velocity:
v^2 = 2*a*s
= 2 * 2.30 * 4.4
= 20.265
thus, v = 4.50 m/s