A body of mass 1 kg undergoes a change of velocity of 4m/s in 4s what is the force acting on it *

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Question 1
How much net force is required to accelerate a 1000 kg car at 4.00 m/s2?
Solution $F=ma$

Given a=4.00 m/s2

m=1000kg Therefore , $F= ma= 1000 \times 4 =4000$ N

Question 2

If you apply a net force of 3 N on .1 kg-box, what is the acceleration of the box

(a) 2 m/s2

(b) 30 m/s2
(c) 10 m/s2 (d) None of these.

Solution

Given F=3 N

m=.1 kg

$F=ma$ or

$a = \frac {F}{m}= 30$ m/s2

Question 3

A body of mass 1 kg undergoes a change of velocity of 4m/s in 4s what is the force acting on it?

Solution

Given $\Delta v=4 m/s$ ,t=4 s ,m=1kg Acceleration is given by $a= \frac {\Delta v}{t}$

a=1 m/s2

Now force is given by

$F=ma$

F=1 N

Question 4.

A particle of 10 kg is moving in a constant acceleration 2m/s2 starting from rest. What is its momentum and velocity per the table given below

S.No	time	Momentum	Velocity
1	1sec
2	1.5 sec
3	2 sec
4	2.5 sec

Solution
Velocity can find using $v=u+at$ For u=0 $v=at$ Momentum $P=mv$

S.No	time	Momentum	Velocity
1	1sec	20 Kg m/s	2 m/
2	1.5 sec	30 kg m/s	3 m/s
3	2 sec	40 kg m/s	4 m/s
4	2.5 sec	50 kg m/s	5 m/s

Question 5 If a net force of 7 N was constantly applied on 400 g object at rest, how long will it take to raise its velocity to 80 m/s? a. 0 s b. 2.23 s c. 3.47 s d. 4.57 s

Solution

Given F=7 N,m=400g= .4 kg Acceleration is given by $a= \frac {F}{m}$

a=17.5 m/s2

Now u=0,v=80 m/s $v=u + at$ $t=\frac {v-u}{a}$

t=4.57 sec

Question 6 A sedan car of mass 200kg is moving with a certain velocity . It is brought to rest by the application of brakes, within a distance of 20m when the average resistance being offered to it is 500N.What was the velocity of the motor car?

Solution

$F=ma$ or $a= \frac {F}{m}$ or

a= -500/200=-2.5 m/s2

Now $v^2=u^2 +2as$

Now v=0,s=20 m,a=-2.5 m/s2

So, u=10 m/s
Question 7
A driver accelerates his car first at the rate of 4 m/s2 and then at the rate of 8 m/s2 .Calculate the ration of the forces exerted by the engines?
Solution $F_1= ma_1$ and $F_2=ma_2$ So,Ratio of force exerted is given by $=\frac {F_1}{F_2}=\frac {ma_1}{ma_2}=\frac {a_1}{a_2}= 1:2$

Question 8

An object of mass 10 g is sliding with a constant velocity of 2 m/ s on a frictionless horizontal table. The force required to keep the object moving with the same velocity is (a) 0 N (b) 5 N (c) 10 N (d)20 N

Solution

As m=0, F=0 Hence (a) is correct

Question 9

A cricket ball of mass 0.20 kg is moving with a velocity of 1.2m/s . Find the impulse on the ball and average force applied by the player if he is able to stop the ball in 0.10s?

Solution

Impulse= Change in momentum $I=\Delta p= m \Delta v= .20 \times 1.2=.12$ Kgm/s Now Impulse is also defined as $I=F \times t$ or $F \times t=.12$ or $F=\frac {.12}{.10}=1.2$ N

Question 10

A car start from rest and acquire a velocity of 54 km/h in 2 sec. Find (i) the acceleration (ii) distance travelled by car assume motion of car is uniform (iii) If the mass of the car is 1000 Kg,what is the force acting on it?

Solution

Given u=0 ,v= 54 km/hr= 15 m/s , t=2 sec a. Acceleration is given by $a=\frac {\Delta v}{t}$

So, a=7.5 m/s2

b. Distance is given by $s=ut+ \frac {1}{2}at^2$ s= 15 m c. Force is given by $F=ma=1000 \times 7.5=7500$ N

Question 11

A hockey ball of mass .2 Kg travelling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 2 m/s . Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution
$ \Delta P= m \times (v - u) = 0.2 \times (-2 - 10) = -2.4$ kg ms-1 (The negative sign indicates a change in direction of hockey ball after it is struck by hockey stick. )

Question 12

Two objects of masses of 100 gm and 200 gm are moving in along the same line and direction with velocities of 2 ms-1 and 1 ms-1 respectively. They collide and after collision, the first object moves at a velocity of 1.67 ms-1. Determine the velocity of the second object?
Solution Given $m_1 = 100 gm = 0.1 kg$, $m_2 = 200 gm = 0.2 kg$

$u_1 = 2$ ms-1, $u_2 = 1$ ms-1, $v_1 = 1.67$ ms-1, $v_2 = ?$

By the law of conservation of momentum, $m_1 u_1+m_2 u_2=m_1 v_2 + m_2 v_2$ $0.1 \times 2+ 0.2 \times 1=0.1 \times 1.67 + 0.2 v_2$

$v_2 = 1.165$ ms-1

It will move in the same direction after collision

Question 13

Anand leaves his house at 8.30 a.m. for his school. The school is 2 km away and classes start at 9.00 a.m. If he walks at a speed of 3 km/h for the first kilometre, at what speed should he walk the second kilometre to reach just in time?

Question 14

An object of mass 1kg acquires a speed of 10 m/s when pushed forward. What is the impulse given to the object?

Solution

Impulse=Change in Momentum= 10 Kgm/s

Question 15

A bullet of mass 10 gm is fired with an initial velocity of 20 m/s from a rifle of mass 4 kg. Calculate the initial recoil velocity of the rifle?

Solution

Let v be the initial recoil velocity of the rifle From law of conservation of Momentum $0=.01 \times 20 + 4 \times v$ Or $v=-.05 m/s$

Question 16

Velocity versus time graph of a ball of mass 100 g rolling on a concrete floor is shown below. Calculate the acceleration and the frictional force of the floor on the ball?

Solution From the graph ,we can see that $\Delta v=-80 m/s$, t=8 sec Now $a= \frac {\Delta v}{t} = -10 m/s^2$ Frictional force will be given as $F= ma= .1 \times -10 = -1 N$

Question 17

Which would require a greater force �� accelerating a 2 kg mass at 5 m/s2 or a 6 kg mass at 2 m/s2?
Solution we have $F = ma$. Here we have $m_1 = 2$ kg and $a_1 = 5 m/s^2$ and $m_2 = 6$ kg and $a_2 = 2 m/s^2$ . So, $F_1 = m_1a_1 = 2 \times 5 = 10 N$ $F_2 = m_2a_2 = 6 \times 2 = 8 N$. Here $F_1 < F_2$

Thus, accelerating a 6 kg mass at 2 m/s2 would require a greater force

Question 17

An object of mass 2 kg is sliding with a constant velocity of 8 m/s on a frictionless horizontal table. The force required to keep the object moving with the same velocity is (a) 16 N (b) 8 N (c) 2 N (d) 0 N

Solution

(d) as there is no friction force

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