Question

The velocity-time graph of a ball moving on the surface of floor is shown in the figure. Calculate the force acting on the ball, if mass of the ball is 100g.

Answer 1

Answer:

• 0.5 Newton is the required Answer.

Explanation:

Given:-

• Initial velocity ,u = 0m/s
• Final Velocity ,v = 20m/s
• Time taken ,t = 4 s
• Mass,m = 100g = 0.1 kg

To Find:-

• Force ,F

Solution:-

Firstly we calculate the acceleration of the body.

Acceleration is defined as the rate of change of velocity at per unit time.

• a = v-u/t

Substitute the value we are

→ a = 20-0/4

→ a = 20/4

→ a = 5m/s².

Acceleration of the body is 5m/s².

Now, Calculating the Force .

Force is defined as the product of mass and Acceleration.

• F = ma

Substitute the value we are

→ F = 0.01×5

→ F = 0.5 N

• Hence, the force acting on the ball is 0.5 Newton.

Answer 2

Answer:

Diagram:-

$\setlength {\unitlength}{1cm}\begin {picture}(20,15)\thicklines\qbezier (0,0)(0,0)(7,0)\qbezier (0,0)(0,0)(0,7)\multiput (1.5,-0.2)(1.5,0){4}{\line (0,1){0.5}}\multiput(-0.2,3)(0,3){2}{\line (1,0){0.5}}\qbezier (0,0)(0,0)(6,6)\thinlines\put (0,6){\line (1,0){1.5}}\put(1.8,6){\line (1,0){1.5}}\put (3.6,6){\line(1,0){2.4}}\put (6,6){\line (0,-1){2.4}}\put (6,3.3){\line (0,-1){1.5}}\put (6,1.5){\line (0,-1){1.5}}\put (-0.2,-0.2){\bf 0}\put (1.5,-0.4){\sf 1}\put (3,-0.4){\sf 2}\put (4.5,-0.4){\sf 3}\put (6,-0.4){\sf 4}\put (-0.6,3){\sf 10}\put (-0.6,6){\sf 20}\put (2,-1){\sf Time(in\: s)}\put(-1.5,2.5){\sf Velocity}\put (-1.8,2){\sf (in\:ms^{-1}) }\put (-0.8,2.8){\vector(0,1){2}}\put (4,-0.9){\vector(1,0){2}}\end {picture}$

Given:-

mass =m=100g=0.01kg

initial velocity =u=0m/s

Final velocity =v=20m/s

time taken=t=4s

To find :-

Force =F

Solution:-

First we have to find acceleration =a

We know that

$\boxed{\sf Acceleration_{(a)}=\dfrac {v-u}{t}}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Acceleration_{(a)}=\dfrac {20-0}{4}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Acceleration_{(a)} =\dfrac{20}{4}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Acceleration _{(a)}=5m/s^2$

_______________________________________

Now according to Newton 's second law of motion

$\boxed{\sf Force_{(F)}=Mass_{(m)}\times Acceleration_{(a)}}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Force=0.01\times 5$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Force=0.5N$

$\\\\\therefore\sf Force\:acting \:on\:the\:ball\:is\:0.5N.$