Question

A body of mass 2 Kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1 . Calculate the (i) work done by the applied force in 10 s (ii) work done by the friction in 10 s *
1 point
(i) 882 J , (ii) 635 J
(i) 635 J (ii) -246.9 J
(i) 882 J (ii) -246.9 J
None of the above​

Answer 1

The mass of the body is m=2kg

The force applied on the body F=7N

The coefficient of kinetic friction =0.1

Since the body starts from rest, the initial velocity of body is zero.

Time at which the work is to be determined is t=10s

The acceleration produced in the body by the applied force is given by Newtons second law of motion as:

$\implies a' = \frac{F}{m} = \frac{7}{2 } = 3.5 \: m /{s}^{2}$

Frictional force is given as:

$\implies F= \mu \: g = 0.1 \times 2 \times 9.8= 1.96 \: N$

The acceleration produced by the frictional force:

$\implies a'' = - \frac{1.96}{2} = - 0.98 \: m /{s}^{2}$

Therefore, the total acceleration of the body:

$\implies a' + a'' = 3.5 + ( - 0.98) = 2.52 \: m {s}^{2}$

The distance traveled by the body is given by the equation of motion:

$\implies \: s = ut + \frac{1}{2} a {t}^{2} \\ \\ \implies \: s = 0 + \frac{1}{2} \times 2.52 \times {(10)}^{2} \\ \\ \implies \: s = 126 \: m$

(a) Work done by the applied force,

$\implies W_a=F.s=7×126=882\:J$

(b) Work done by the frictional force,

$\implies W_f=F.s=1.96×126=247\:J$

work done by friction force is opposite to work done by applied force

i.e $W_f=-247\:J$

Hence, option c) i) 882 J ii) -246.99 J is the correct option.

Answer 2

Explanation:

$\huge{\underline{\underline{\boxed{\sf{Question}}}}}$

A body of mass 2 Kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1 . Calculate the (i) work done by the applied force in 10 s (ii) work done by the friction in 10 s *

1 point

(i) 882 J , (ii) 635 J

(i) 635 J (ii) -246.9 J

(i) 882 J (ii) -246.9 J

None of the above

$\huge{\underline{\underline{\boxed{\sf{Answer}}}}}$

The mass of the body is m=2kg

The force applied on the body F=7N

The coefficient of kinetic friction =0.1

Since the body starts from rest, the initial velocity of body is zero.

Time at which the work is to be determined is t=10s

The acceleration produced in the body by the applied force is given by Newtons second law of motion as:

$\implies a' = \frac{F}{m} = \frac{7}{2 } = 3.5 \: m /{s}^{2}$

Frictional force is given as:

$\implies F= \mu \: g = 0.1 \times 2 \times 9.8= 1.96 \: N$

The acceleration produced by the frictional force:

$\implies a'' = - \frac{1.96}{2} = - 0.98 \: m /{s}^{2}$

Therefore, the total acceleration of the body:

$\implies a' + a'' = 3.5 + ( - 0.98) = 2.52 \: m {s}^{2}$

The distance traveled by the body is given by the equation of motion:

$\implies \: s = ut + \frac{1}{2} a {t}^{2} \\ \\ \implies \: s = 0 + \frac{1}{2} \times 2.52 \times {(10)}^{2} \\ \\ \implies \: s = 126 \: m$

(a) Work done by the applied force,

$\implies W_a=F.s=7×126=882\:J$

(b) Work done by the frictional force,

$\implies W_f=F.s=1.96×126=247\:J$

work done by friction force is opposite to work done by applied force

i.e $W_f=-247\:J$

Hence, option c) i) 882 J ii) -246.99 J is the correct option.