Question

A 3.0-kg block is dragged over a rough horizontal surface by a constant force of 16 N acting at an angle of 37° above the horizontal as shown. The speed of the block increases from 2 m/s to 3.8 m/s in a displacement of 5.5 m. What work was done by the friction force during this displacement?

Answer 1

The value is
`W = -54.615 \ J`

Step-by-step explanation:

From the question we are told that

The mass of the block is
`m = 3.0 \ kg`

The force is
`F = 16 \ N`

The angle is
`\theta = 37^o`

The first speed of the block is
`u = 2 \ m/s`

The second speed of the block is
`v = 3.8 \ m/s`

The displacement is
`d = 5.5 \ m`

Gnerally from kinematic equation we have that

`v^2 = u^2 + 2as`

=>
`3.8 ^2 = 2^2 + 2 * a* 5.5`

=>
`a = 0.9491 \ m/s^2`

Generally the net force acting on the crate is mathematically represented as

`F_(net) = [ F cos (\theta ) - F_f ] = ma`

Here
`F_f` is the frictional force acting on the crate

So

`[ 16 cos (37 ) - F_f ] = 3 * 0.9491`

=>
`F_f = 9.93 \ N`

Generally the work done by friction during the displacement is mathematically represented as

`W = F_f * d * cos ( 180 )`

=>
`W = 9.93 * 5.5 * cos ( 180 )`

=>
`W = -54.615 \ J`