Question

Q33 A body of 1 kg mass is pushed
uphill against gravity and frictional
force to a height of 10 m (top of the
cliff) slowly and then turned and
dropped with a velocity of 2 m/s
downhill, if the body stops at a
height of 6 m and 3 m horizontally
away from the cliff's summit, the
frictional force acting on the body
is:​

Answer 1

Answer:

The force action on the body is 0.50 N

Step by step solution

Given that,

Speed $=2 \mathrm{~m} / \mathrm{s}$

Height $=10 \mathrm{~m}$

Mass $=1 \mathrm{~kg}$

According to the figure,

We need to calculate the distance

Using Pythagorean theorem

$A C=\sqrt{A B^{2}+B C^{2}}$

Put the value into the formula

$\begin{aligned}&A C=\sqrt{6^{2}+3^{2}} \\&A C=6.7 \mathrm{~m}\end{aligned}$

We need to calculate the acceleration

Using equation of motion

$v^{2}=u^{2}+2 a s$

Put the value into the formula

$\begin{aligned}&0=2^{2}+2 \times a \times 6.7 \\&a=-\frac{4}{6.7} \\&a=-0.59 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}$

Negative sign shows the deceleration.

We need to calculate the force action on the body

Using formula of force

$F=m a$

Put the value into the formula

$\begin{aligned}&F=1 \times 0.59 \\&F=0.50 \mathrm{~N}\end{aligned}$

Hence, The force action on the body is 0.50 N