1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A force of 5 kg – wt is acting on a body of mass 5 kg kept on a smooth horizontal surface at rest. If it acts for 10 seconds at an angle of 60∘ with the horizontal, find the distance travelled by it along the surface (Take g=10 m/s2)

A
100 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
150 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
200 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
250 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 250 mWe know that 1 Kg - wt=g N=10 N Hence, F=5×10 N=50 N The component of the force along the horizontal surface =Fs=50cos60∘=25 N Hence, acceleration of the body along the surface, a=Fsm=255=5 m/s2 Distance travelled by the body along the surface in 10 s, s=ut+12at2=0+12×5×100=250 m

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program