Splitting weight on a slope

Mathematics · Mechanics · ages 17-18

Name ______________________ Date ____________

A 2 kg block slides on a fixed smooth slope. Its weight components are 12 N downhill and 16 N into the slope. Only weight and the normal reaction act. What is its downhill acceleration in m/s squared?
Answer: ______________
A block slides on a fixed smooth slope at angle theta. Only weight and the normal reaction act. Choose uphill positive. Which equation gives its signed acceleration a along the slope?
- mg cos(theta) = ma
- -mg sin(theta) = ma
- mg sin(theta) = ma
- -mg cos(theta) = ma
A block slides on a smooth slope at 30 degrees above horizontal. Which component of its weight acts down the slope?
- mg sin30 degrees
- mg cos30 degrees
- mg tan30 degrees
A block stays on a fixed straight smooth slope with angle 0 < theta < 90 degrees. Only weight and the normal reaction act. Its normal reaction equals the full weight mg.
Circle one: True False
An 18 N rope pulls a 3 kg body at 60 degrees above horizontal on a smooth level floor. Only rope, weight and reaction act. Take g = 10 m/s squared. What is its forward acceleration in m/s squared?
Answer: ______________
Two identical blocks slide on fixed smooth slopes at 30 and 60 degrees, with the same g. Only weight and the normal reaction act on each. Which has the greater downhill acceleration?
- the 30 degree slope
- the 60 degree slope
- both give 4.9 m/s squared
A 2 kg block slides on a fixed smooth 60 degree slope. An extra 8 N force presses perpendicular into the slope. Only this force, weight and the reaction act. Take g = 10 m/s squared. What is R?
- 10 N
- 28 N
- 18 N
- 12 N
Only rope, weight and reaction act on a body on a smooth level floor. Keep pull magnitude and mass fixed, and keep contact. Increasing the rope angle above horizontal from 0 toward 90 degrees reduces forward acceleration.
Circle one: True False
A block slides on a fixed straight smooth 60 degree slope. Only weight and the normal reaction act. Its normal reaction is 24 N. Take g = 10 m/s squared. What is its mass in kilograms?
Answer: ______________
A 3 kg block on a fixed smooth 30 degree slope has a 24 N rope pulling at 60 degrees above uphill. Only rope, weight and reaction act. Take g = 10 m/s squared; contact remains. What is its acceleration?
- 4 m/s squared uphill
- 5 m/s squared downhill
- 9 m/s squared uphill
- 1 m/s squared downhill

Answer key

For grown-ups. Fold this page away before handing over the rest.

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6 · The reaction balances the perpendicular 16 N component. The downhill resultant is 12 N, so a = 12 / 2 = 6 m/s squared.
-mg sin(theta) = ma · The parallel weight component points downhill, opposite the positive axis. Its signed value is -mg sin(theta).
mg sin30 degrees · The downhill component is mg sin30 degrees. The component perpendicular into the slope is mg cos30 degrees.
False · The reaction balances mg cos(theta). In the stated angle range, cos(theta) < 1, so R < mg.
3 · The forward resultant is 18 cos60 degrees = 9 N. Hence a = 9 / 3 = 3 m/s squared.
the 60 degree slope · Each acceleration is g sin(theta). Since sin60 degrees exceeds sin30 degrees, the 60 degree slope gives greater acceleration.
18 N · Weight contributes 2 × 10 × 0.5 = 10 N into the slope. The extra inward force adds 8 N, so R = 18 N.
True · Forward acceleration is F cos(alpha)/m. With F and m fixed, it decreases as alpha increases through this range.
4.8 · Use 24 = m × 10 × 0.5. Thus 24 = 5m, giving m = 4.8 kg.
1 m/s squared downhill · The uphill rope component is 12 N; the downhill weight component is 15 N. The resultant is 3 N downhill, giving 1 m/s squared downhill.