Question

Joe pushes down the length of the handle of a 10.9 kg lawn spreader. The handle makes an angle of 45.3 ◦ with the horizontal. Joe wishes to accelerate the spreader from rest to 1.35 m/s in 1.6 s. What force must Joe apply to the handle? Answer in units of N

Answer 1

Answer:

F = 13.08 N

Explanation:

given,

mass of spreader = 10.9 Kg

angle made with angle = 45.3°

accelerate the spreader to 1.35 m/s in 1.6 s.

Acceleration = $\dfrac{v}{t}$

                      = $\dfrac{1.35}{1.6}$

                      = 0.844 m/s²

momentum in = momentum out

F × t = m  × v

$F cos \theta \times t = m \times v$

$F cos \ 45.3^0 \times 1.6 = 10.9 \times 1.35$

F×1.125 = 14.175

$F = \dfrac{14.175}{1.125}$

F = 13.08 N

hence, the force applied by the joe is equal to F = 13.08 N