Question

a 5 kg block is kept on a horizontal platform at rest at time zero the platform start moving with constant acceleration 1 metre per second square the coefficient of friction between the block and the platform is 0.2 the work done by the force of friction on the block in the fix reference frame in 10 second is​

Answer 1

Answer:

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Answer 2

Given that,

Mass of block = 5 kg

Acceleration = 1 m/s²

Coefficient of friction = 0.2

Time = 10 sec

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times1\times10^2$

$s=50\ m$

We need to calculate the friction  force

Using formula of friction

$f_{f}=\mu mg$

Put the value into the formula

$F_{f}= 0.2\times5\times10$

$F_{f}=10\ N$

We need to calculate the force due to acceleration

Using formula of force

$F=ma$

Put the value into the formula

$F=5\times1$

$F=5\ N$

Friction is self adjusting so the friction force is 5 N.

Frictional force and displacement are in opposite direction.

We need to calculate the work done by the force of friction on the block in the fix reference frame

Using formula of work done

$W=F_{f}\cdot s\cos\theta$

Put the value into the formula

$W=5\times50\cos180$

$W=-250\ J$

Negative sign show the opposite direction of force.

Hence, The work done by the force of friction on the block is 250 J.