Question

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 cliffs summit, the frictional force acting on the body is:

Answer 1

Given that,

Speed = 2 m/s

Height = 10 m

Mass = 1 kg

According to figure,

We need to calculate the distance

Using pythagorean theorem

$AC=\sqrt{AB^2+BC^2}$

Put the value into the formula

$AC=\sqrt{6^2+3^2}$

$AC=6.7\ m$

We need to calculate the acceleration

Using equation of motion

$v^2=u^2+2as$

Put the value into the formula

$0=2^2+2\times a\times 6.7$

$a=-\dfrac{4}{6.7}$

$a=-0.59\ m/s^2$

Negative sign shows the deceleration.

We need to calculate the force action on the body

Using formula of force

$F=ma$

Put the value into the formula

$F=1\times0.59$

$F=0.50\ N$

Hence, The force action on the body is 0.50 N

Learn more :

Topic : kinematics

https://brainly.in/question/18327919

Answer 2

Answer:

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