Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A force of 50N acts in the direction as shown in fig. The block of mass of 5kg resting on a smooth horizontal surface . What is the acceleration of the block? <img alt="" src="https://cdn

A force of 50N acts in the direction as shown in fig. The block of mass of 5kg resting on a smooth horizontal surface . What is the acceleration of the block?

Option: 1

5m/s^2


Option: 2

8m/s^2


Option: 3

5\sqrt3m/s^2


Option: 4

10m/s^2


Answers (1)

Solution :

\\ \text{Free body diagram of given system :}

\\ \text{Horizontal component of the force }= F\sin\theta=50\sin60^{\circ}=\frac{50\sqrt{3}}{2}N

\\ \text{acceleration of the block,(a)}=\frac{F\sin\theta}{M}=\frac{\frac{50\sqrt{3}}{2}}{5}=5\sqrt{3} \ m/s^{2}

Posted by

Kshitij

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions