You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Physics

Work

A body of mas...

Question

A body of mass $2 k g$ initially at rest moves under the action of an applied horizontal force of $7 N$ on a table with coefficient of kinetic friction $= 0.1$ . Compute the Work done by the applied force in $10 s$

875 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

890 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1000 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5000 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $875 J$
Here, u=0(initial speed)
Friction force=0.1210=2N
Net force on the body=7N-2N=5N
So, $a = 5 / 2 k g = 2.5 m / s^{2}$
Therefore, distance travelled in 10 sec = $1 / 2 a t^{2}$
= $1 / 2 * 2.5 * 100$
=125 m
Now, work done by the applied force=F.S =125*7=875 J

Suggest Corrections

Similar questions

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.

Compute the

(a) work done by the applied force in 10 s,

(b) work done by friction in 10 s,

(d) change in kinetic energy of the body in 10 s,

and interpret your results.

Q. 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

μ_{k} = 0.1

.
Calculate the work done by kinetic friction in

10 sec (g = 9.8 {m/s}^{2})

Q. 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. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results.

Q. 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

μ_{k} = 0.1

.
Calculate the work done by kinetic friction in

10 sec (g = 9.8 {m/s}^{2})

Q. A block of mass

2 k g

initially at rest moves under the action of an applied horizontal force of

6 N

on a rough horizontal surface. The coefficient of friction between block and surface is

0.1

. The work done by the applied force in

10 s

is (Take

g = 10 m s^{- 2})

Join BYJU'S Learning Program

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. Compute the Work done by the applied force in 10 s

The correct option is A 875 JHere, u=0(initial speed)Friction force=0.1*2*10=2N Net force on the body=7N-2N=5NSo,a=5/2kg=2.5m/s2Therefore, distance travelled in 10 sec =1/2at2 =1/2∗2.5∗100 =125 mNow, work done by the applied force=F.S =125*7=875 J

A body of mass $2 k g$ initially at rest moves under the action of an applied horizontal force of $7 N$ on a table with coefficient of kinetic friction $= 0.1$ . Compute the Work done by the applied force in $10 s$

The correct option is A $875 J$
Here, u=0(initial speed)
Friction force=0.1210=2N
Net force on the body=7N-2N=5N
So, $a = 5 / 2 k g = 2.5 m / s^{2}$
Therefore, distance travelled in 10 sec = $1 / 2 a t^{2}$
= $1 / 2 * 2.5 * 100$
=125 m
Now, work done by the applied force=F.S =125*7=875 J