Question

A block of mass 5 kg is placed on horizontal surface and a pushing force 20 N is acting on back as
shown in figure. If coefficient of friction between block and surface is 0.2, then calculate frictional force
and speed of block after 15 S. (Given g = 10 m/s2)
20 N
45°
(1) 2.936 MS-1
(2) 4.936 MS-1
(3) 3.936 MS-1
(4) None of these​

Answer 1

$v=12.426\ m.s^{-1}$ is the velocity after 15 seconds.

Explanation:

Given:

• mass of block, $m=5\ kg$

• pushing force on the block, $F=20\ N$

• coefficient of friction between the contact surfaces, $\mu=0.2$

• time, $t=15\ s$

Assumptions:

The force acts at an angle of 45° to the displacement of block.

Normal reaction force on the block will be equal to the weight of the block.

$F_R=w$

$F_R=m.g$

$F_R=5\times 10$

$F_R=50\ N$

Now the frictional force:

$f=\mu.F_R$

$f=0.2\times 50$

$f=10\ N$    (opposite to the direction of relative motion)

Now the net force acting on the body in the direction of motion:

$F_n=20\ cos 45^{\circ}-10$

$F_n=4.142\ N$

∴Acceleration:

$a=\frac{F_n}{m}$

$a=\frac{4.142}{5}$

$a=0.8284\ m.s^{-2}$

Since the bock was placed on the table, initially it was at rest.

Using the equation of motion:

$v=u+at$

$v=0+0.8284\times 15$

$v=12.426\ m.s^{-1}$

TOPIC: Work, friction, equation of motion.

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