Joe pushes down the length of the handle of a 10.9 kg lawn spreader. The handle

Question

Joe pushes down the length of the handle of
a 10.9 kg lawn spreader. The handle makes
an angle of 42.7 with the horizontal. Joe
wishes to accelerate the spreader from rest to
1.38 m/s in 1.5 s.
What force must Joe apply to the handle?
Answer in units of N.

My approach:
1) vf=vo+at
a=(1.38m/s)/(1.5s)=.92 m/s^2

2) sum of Fx= Fcos42.7=ma
F=(10kg)(.92m/s^2)/(cos42.7)=12.52N
Where is my mistake? I got the question wrong. Do I need to plug 12.52N into Fcos42.7 and that will be my answer?

Please advise. Thanks!

Explanation / Answer

Momentum in = Momentum out F x t = m x v Fcos(theta) x t = m x v Fcos(42.7) x 1.5 = 10.9(1.38) F = 13.6451 = 13.6 N Hope this helps! Rate lifesaver :)

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