1. A water skier has a mass of 79 kg and accelerates at 1.4 m/s2. What is the net force acting on Him?
2. What is the mass of an object if it takes a net force of 32N to accelerate it at a rate of 0.88 m/s2?
3. A net force of 15N is applied to a cart with a mass of 2.1 kg.
a. What is the acceleration of the cart?
b. How long will it take the cart to travel 2.8 m, starting from rest?
4. What is the acceleration of a box weighing 666 N if a force of 777 N is applied to it?
5. A car has a mass of 820 kg. It starts from rest and travels 41 m in 3.0 seconds. What is the net force applied to the car?
Answers
Answer:
Practice Problems 1. A water skier has a mass of 79 kg and accelerates at 1.4 m/s2. What is the net force acting on him? m=79 kg a=1.4 m/s2 FnetF=ma F=79x1.
Answers:
1 - 110.6 N
2 - 36.36 kg
3 - (a)- 7.14 m/s²
(b)- 0.88 sec
4 - 11.43 m/s²
5 - 7470.2 N
Step by step solution:
Step 1: Required Knowledge
Newton's second law of motion states the formula -
F = Ma
where, F is the force on the object, M is the mass of the object, a is the acceleration of object.
Step 2: Solving 1
Given data, m = 79 kg, a = 1.4 m/s²
Putting values in the equation of step 1,
F = 79×1.4
= 110.6 N
Step 3: Solving 2
Given data, F = 32 N , a = 0.88 m/s²
Putting values in the equation of step 1,
32 = m×0.88
⇒ m = 32/0.88
⇒ m = 36.36 kg
Step 4: Solving 3
Given data, F= 15 N, m = 2.1 kg
Putting values in the equation of step 1,
(a) 15 = 2.1×a
⇒ a = 7.14 m/s²
(b) Given, u= 0 m/s, s = 2.8 m
Putting the values,
⇒ t = 0.88 sec
Step 5: Solving 4
Given data, weight = 666 kg-wt , F = 777 N
mass of box= 666/9.8 = 67.95 kg (∵ weight= mg)
Now, putting values in equation of step 1,
777 = 67.95×a
⇒ a= 11.43 m/s²
Step 6: Solving 5
Given data, m = 820 kg, u=0 m/s , d= 41 m, t= 3.0 sec
Putting values,
41 = 0 + (1/2)×a×3²
⇒ a = 9.11 m/s²
Now, F = 820× 9.11
= 7470.2 N