Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

Consider a block of mass 5 kg on a table top initially at rest. Assume that the surface is frictionless. Let a pushing force of 50 N directed at an angle of 30$^\circ$ below the horizontal act on the block. Find the acceleration of the block and the normal force that the table exerts on the block.

Can you answer this question?

1 Answer

0 votes
Answer: 8.66 ms$^{-2}$ and 74 N
Here is the force diagram for the problem.
According to Newton's 2nd law, $F = ma_x = F_P \cos 30^\circ$
Given that $F_P = 50N,\;$ and $m = 5\; Kg \rightarrow a_x = \large\frac{50N \times \cos 30 ^\circ}{5 kg}$$ = 8.66 m/s^2$
From the diagram above, we see that $N - mg - F_P \sin30^\circ = 0$
$\Rightarrow N = mg + F_P \sin 30^\circ = 5 \times 9.8 + 50 \sin 30 = 74 \;N$
answered Aug 19, 2014 by balaji.thirumalai
edited Aug 19, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App