PS-10-Force Problems worksheet - Free Printable

Let's solve each problem step by step using Newton’s Second Law of Motion:

> F = m × a
> Where:
> - F = Force (in Newtons, N)
> - m = mass (in kg)
> - a = acceleration (in m/s²)

We'll use the triangle method as shown in the worksheet:
- If you cover F, then F = m × a
- If you cover m, then m = F ÷ a
- If you cover a, then a = F ÷ m

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1. A 40-kg baseball is accelerated at 600 m/s² by what force?

Given:
- m = 40 kg
- a = 600 m/s²

Use:
F = m × a
F = 40 × 600 = 24000 N

✔ Answer: 24000

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2. The acceleration of gravity is 9.8 m/s² on Earth. An object has a mass of 700 kg. What is its weight?

Weight = Force of gravity = m × g
- m = 700 kg
- g = 9.8 m/s²

F = 700 × 9.8 = 6860 N

✔ Answer: 6860

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3. The weight of an object on a distant planet is 600 N, and its mass is 294.56 kg. What is the acceleration of gravity on that planet?

Given:
- F = 600 N
- m = 294.56 kg

Use:
a = F ÷ m
a = 600 ÷ 294.56 ≈ 2.04 m/s²

✔ Answer: 2.04

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4. A weightlifter raises a 500 kg barbell with an acceleration of 3 m/s². How much force does the weightlifter use?

The total force must overcome gravity and provide upward acceleration.

So, net force = m × a_total
But we need to consider that the barbell already has gravitational force acting downward.

However, the question asks for the force the weightlifter applies, which must counteract gravity and provide extra force for acceleration.

So:
- Weight (gravity) = m × g = 500 × 9.8 = 4900 N
- Additional force for acceleration = m × a = 500 × 3 = 1500 N
- Total applied force = 4900 + 1500 = 6400 N

Alternatively:
F_net = m × a → but this is only the *net* force.
To get the applied force, we use:

F_applied = m × (g + a)
= 500 × (9.8 + 3) = 500 × 12.8 = 6400 N

✔ Answer: 6400

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5. If a 50 kg water rocket takes off with a force of 250 N, what is its acceleration?

Given:
- F = 250 N
- m = 50 kg

Use:
a = F ÷ m
a = 250 ÷ 50 = 5.00 m/s²

✔ Answer: 5.00

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6. A dancer lifts his partner above his head with an acceleration of 3 m/s². The dancer exerts a force of 175 N. What is the mass of his partner?

Given:
- F = 175 N
- a = 3 m/s²

But again, this force includes overcoming gravity and accelerating upward.

So, the net force is:
F_net = m × a
But the applied force must be:
F_applied = m × (g + a)
→ Because the dancer must support weight (mg) and accelerate upward (ma)

So:
F = m × (g + a)
175 = m × (9.8 + 3) = m × 12.8

Now solve for m:
m = 175 ÷ 12.8 ≈ 13.67 kg

✔ Answer: 13.67

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✔ Final Answers:

1. 24000
2. 6860
3. 2.04
4. 6400
5. 5.00
6. 13.67

All answers are in correct units (N or m/s²), and decimals are used where needed (2 decimal places if not whole). No commas.