A 32,5kg wagon is towed up a hill inclined at 18.9 degrees with respect to the horizontal. The tow rope is parallel to the incline and exerts a force of 117 N on the wagon. Assume that the wagon starts from rest at the bottom of the hill and disregard friction. The acceleration of gravity is 9.81m/s^2. how fast is the wagon going after moving 54.7m up the hill? Answer in units of m/s.
The tow rope part is just really throwing me off. Can someone maybe explain it?
The tow rope part is just really throwing me off. Can someone maybe explain it?
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There are 2 forces parallel to the plane. The resultant of these 2 forces causes the acceleration along the plne. The 2 forces are:
- the tension of 117N upwards parallel to the plane and
- the component of the wagon's weight downwards parallel to the plane ( = mg(sin(θ) ) .
If you don't know where the 'mgsin(θ)' comes from, watch the last part of the video in the link (which explains how to resolve a vector into components).
The resultant force parallel to the pane = 117 - mg(sin(θ)
= 117 - 32.5x9.81xsin(18.9º)
= 13.727N
Using F = ma gives a = F/m
a = 13.727/32.5
= 0.42237m/s²
Using v² = u² + 2as (or vf² = vi² + 2ax if you use those symbols)
v² = 0² + 2x0.42237x54.7
=46.21
v = 6.80m/s² (give answer to 3 significant figures, same as the data supplied)
- the tension of 117N upwards parallel to the plane and
- the component of the wagon's weight downwards parallel to the plane ( = mg(sin(θ) ) .
If you don't know where the 'mgsin(θ)' comes from, watch the last part of the video in the link (which explains how to resolve a vector into components).
The resultant force parallel to the pane = 117 - mg(sin(θ)
= 117 - 32.5x9.81xsin(18.9º)
= 13.727N
Using F = ma gives a = F/m
a = 13.727/32.5
= 0.42237m/s²
Using v² = u² + 2as (or vf² = vi² + 2ax if you use those symbols)
v² = 0² + 2x0.42237x54.7
=46.21
v = 6.80m/s² (give answer to 3 significant figures, same as the data supplied)