Question

A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?

Answer 1

15.25 N

Step-by-step explanation:

A force of
`55\text{ }N` is acting on a wagon along the road. The wagon weights
`7.5\text{ }kg`. Acceleration of the wagon is given as
`5.3\text{ }(m)/(s^(2))`.

Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

Gravitational Force and Normal Reaction cancel out each other.

Net External Force = Mass of system/wagon
`*` Acceleration of wagon

`F_(ext)-F_(friction)=(7.5\text{ }kg)*(5.3\text{ }(m)/(s^(2)))=39.75\text{ }N\\55\text{ }N-F_(friction)=39.75\text{ }N\\F_(friction)=15.25\text{ }N`

`F_(friction)` has a negative sign because it opposes the motion of the wagon.

∴ Frictional Force = 15.25 N