A 292 kg motorcycle is accelerating up along a ramp that is inclined 30.0° above the horizontal. The propulsion force pushing the motorcycle up the ra

Falak Kinney 2020-11-05 Answered
A 292 kg motorcycle is accelerating up along a ramp that is inclined 30.0° above the horizontal. The propulsion force pushing the motorcycle up the rampis 3020 N, and air resistance produces a force of 250 N that opposes the motion. Find the magnitude of the motorcycles
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Expert Answer

Laaibah Pitt
Answered 2020-11-06 Author has 98 answers

What are the sum of forces on the motorcycle? Theproblem states the following:
The motorcycle is pushed up the ramp with force 3020 N
Air resists the motorcyles forward motion, therefore it pusheswith a force of 250 N down the ramp.
The only other force is that due to the motorcycles weight(a. force of gravity). Then the force diagram looks like);
image
Where;
F1=3020 Newtons
F3=250 Newtons
F4 is force due to gravity and F2 is the component of that force parallel to the ramp.
So, F2=F4sin30 or mgsin30
F2=(292)(9.8)sin(30)=1430.8
now the sum of the forces is equal to the mass times acceleration of the motorcycle.
F=F1F2F3 or 30202501430.8=1339.2 Newns
Then the acceleration of the motorcycle is:
F=ma  or  a=Fm=1339.2292=4.586ms2 up the slope
The second part of the problem is also done by summing forces. The force of the block on the surface is due to the mass of the block and mass of the person standing on it. Thisforce must be exactly opposed by the surface (since the block is not moving in the vertical direction).
Then the normal force is:
Fn=(m1+m2)g; where g is the acceleration to gravity.
Fn=(41+69)9.8=1078 Newtons, perpendicular to thesurface and pointing up
Similarly, the person applies a force to the block surface (due to weight) and the block surface response with equal and opposite force:
Fn=699.8=676.2 Newtons, perpendicular to block surfaceand pointing up.

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