Honors Physics Name:
5.4 Incline Plane Review Problems
Draw a free body diagram for the following situations.
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3. A 100 kg box is sliding down an incline at a constant speed of 0.5 m/s. The incline angle
is 25⁰ from the horizontal. Draw a free body diagram and calculate the magnitude of the
normal force.
Need to calculate weight. W=mg = 100 x 9.8 = 980N
F =W x cos ϴ
F = 980 x cos 25°
F = 888 N
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4. A 215 kg box is placed on an inclined plane that makes a 35⁰ angle with the horizon. Find
the component of the weight that is parallel to the surface of the incline.
Need to calculate weight. W=mg = 215 x 9.8 = 2107N
The parallel component of the weight is: W x sin ϴ = 2107N x sin 35° = 1208.5 N
5. A 62 kg person on skis is going down a hill sloped at 37⁰. The coefficient of friction
between the skis and the snow is 0.15. How fast is the skier going 5 sec after starting from
rest?
1st step: Calculate weight. W=mg = 62 x 9.8 = 607.6N
2nd: Find the normal force. On on inclined plane, it is the perpendicular component of the
weight or, W x cos ϴ
F = 607.6 x cos 37° = 485N
3rd: The normal force can be used to calculate the friction force by f =µ F
f =0.15 x 485N = 72.8 N
4th: The downhill force is equal to the parallel component of the weight or, W x sin ϴ =
F = 607.6N x sin 37° = 365.7N
5th: Now the net downhill force can be calculated:
F = F -f
F = 365.7N - 72.8N = 292.9N
6th: If the net force is known, then the acceleration can be calculated, by N2, F =ma
292.9 N = 62 kg x a
a = 4.73 m/s
7th: Nowing acceleration, the velocity can be solved for with K1: v= v + at
v= 4.73 x 5 = 23.7 m/s
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6. You push a 325N trunk up a 20⁰ inclined plane at a constant velocity by exerting a 211N
force parallel to the plane’s surface.
a) what is the component of the trunk’s weight parallell to the plane?
b) what is the sum of all the forces parallel to the planes surface?
c) what is the magnitude and direction of the friction force?
c) what is the coefficent of friction?
a) W x sin ϴ = 325N x sin20° = 111 N
b) since the object is traveling at a constant velocity, the sum of the forces are zero
c) Since the kinetic friction force needs to balance the component of the weight, the
magnitude is equal to part (a), the direction is opposite. -111 N
d) Need to calculate the normal force, W x cos ϴ = 325 x cos20° = 305.4N
f =µ F
111 N =µ x 305.4N
µ = 0.36
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7. A 75 kg skier speeds down a trail, the surface is smooth and inclined at an angle of 22⁰
with the horizontal.
a) Find the direction and magnitude of the net force acting on the skier.
b) Does the net force exerted on the skier increase, decrease, or stay the same as
the slope becomes steeper?
a) Ignoring friction, the net force acting on the skier is the parallel component ( downhill
component) of the weight.
W=mg = 75 x 9.8 = 735 N
F = W x sin ϴ = 735N x sin22° = 275 N
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b) as the slope becomes steeper the force on the skier increases. More of the component of
the weight is shifted downhill. A 90° slope would have 100% of the weight (sin 90° = 1), or
the skier would be in a vertical freefall.