# Question #e8125

Jan 11, 2018

Normal force= 24 N
Frictional force acting downwards of 9.6 N in magnitude
Acceleration of the system is 1.2 $\frac{m}{s} ^ 2$

#### Explanation:

Normal force acting is given by N=F $\sin 37$ i.e the force applied horizontally on the block.
N= $40 \cdot \left(\frac{3}{5}\right)$ or 24 N
As the system is being mobilised upward frictional force will be acting against the direction of effective motion.
Its value will be $u \cdot N$
Or,$\left(0. 40 \cdot 24\right)$N i.e 9.6 N
Now,let's consider that the system moves upward with acceleration a,so from equation of force we can write,
F $\cos 37 - \left(u \cdot N + m \cdot g\right)$=$m \cdot a$
Or,$a = 1.2 \frac{m}{s} ^ 2$