Practice on Force, Weight, Friction, and Acceleration: Solution-1
Question 1: If your hand exerts a force of 14 N vertically (upward) on a box weighting 13 N.
What would be the reaction to hand's force on the box?
Answer: 14 N
Explanation: the reaction to hand's force on the box is equal to your hand's force.
Question 2: Assume that the surface gravity of Mars is 4 m/s2 and an astronaut with mass of
80 kg were parachuting on this planet.
Calculate the force exerted by air resistance at terminal velocity (i.e., gravity or weight is equal to the opposite force of air resistance).
Answer: 320 N
Explanation: force magnitude = surface gravity of Mars × mass of the astronaut = 4 m/s2 × 80 kg = 320 N.
Question 3: The net force acting on a body gives the direction of the ________ of the body.
Question 4: Assume a body of mass 1 kg is pushed across a horizontal table by a force of 1 N and
the acceleration was measured at 0.8 m/s2.
Calculate the force of friction.
Answer: 0.2 N
Exlanation: F net= acceleration (a) × mass (m) = 0.8 m/s2 × 1kg = 0.8 N.
Hence the friction = F total - F net = 1 N = 0.8 N = 0.2 N
Question 5: A block of mass 5 kg is acted upon by a single force, generating an acceleration of 3.0 /s2.
The value of the force is
Answer: 15 N
Exlanation: F net= acceleration (a) × mass (m) = 3.0 m/s2 × 5kg = 15 N.
Question 6: A block of mass 20 kg on horizontal surface without friction is acted upon by two forces:
a horizontal force of 90 N acting to the right and a horizontal force of 50 N to the left.
Find the acceleration of the block and its direction.
Answer: 2.0 m/s2.
- 1 m/s2
- 2 m/s2
- 5 m/s2
- 4 m/s2
Explanation: F net= 90 N − 50 N = 40 N to the right and F net = acceleration (a) × mass (m), so a = 40 N / 20 kg = 2 m/s2
Question 7: Assume a crate is acted upon by a net force of 200 N and an acceleration is measured at 5.0 m/s2.
Calculate the weight of the crate.
Answer: 392 N
Explanation: F = mass (m) × acceleration (a), which implies that m = 200 N / 5 m/s2 = 40 kg.
Weight = mass × 9.8 m/s2 = 40 × 9.8 m/s2 = 392 N
Question 8: A block of mass 100 kg that is free to roll on a level surface is acted upon a net horizontal force of 400 N.
The acceleration is
Answer: 4 m/s2.
- 1 m/s2
- 2 m/s2
- 3 m/s2
- 4 m/s2
Explanation: F = mass (m) × acceleration (a), and so a = 400 N / 100 kg = 4 m/s2
Question 9: The frictional force, due to air resistance, acting on an object is always
Answer: in the opposite direction to the object's motion.
- in the opposite direction to the object's friction.
- in the opposite direction to the object's motion.
- in the opposite direction to the object's mass.
- in the opposite direction to the object's vertical force.
Question 10: Two blocks B1 and B2 having identical masses but the net force applied to B1 is 50 N and to B2 is 100 N.
What is the relationship between the two accelerations?
Answer: Acceleration of B2 is larger.
- Acceleration of B2 is larger
- Acceleration of B1 is larger.
- no relationship
Explanation: F1 = mass of B1 (m) × acceleration (a1) = 50 N, and F2 = mass of B2 (m) × acceleration (a2) = 100 N
Using using mathematical substitution, we obtain from the first equation m = 50 / (a1) and then substituting this in equation 2, we get 100 N = (50 / a1) × (a2)
So, 50 × a2 = 100 × a1 , and then we obtain a2 = 2 × a1.
Question 11: Assume your weight is 700 newtons and you picked up a box from the ground with a force of 100 newtons.
Find the force the box exerts on you.
Answer: 100 N
Explanation: The force the box exerts on you is equal to the force used to picked up the box from the ground.
Question 12: Assume a force is exerted on person P1 having a mass of 70 kg generating an acceleration of 2.0 m/s2.
The same force is being used on 60-kg person 2 would cause
Answer: a larger acceleration.
- a smaller acceleration.
- a larger acceleration.
- same acceleration.
- zero acceleration.
Explanation: The force exerted on P1 = The force exerted on P2
Hence, mass of P1 (m1) × acceleration (a1) = mass of P2 (m2) × acceleration (a2)
which means, 70 kg × 2.0 m/s2 = 60 kg × a2, and using mathematical substitution, we obtain a2 = 140 / 60 = 2.33 m/s2
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Date of last modification: Summer , 2019