Question

If the sled is also being acted on by a constant kinetic friction force of 260 N and it started from rest, what is the final speed of the sled

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

`W = 121271.99 \ J`

b

`v = 26.74 \ m/s`

Step-by-step explanation:

From the question we are said that

The mass of the shed is
`m_s = 375 \ kg`

The pulling force is
`F = 2511 \ N`

The angle is
`\theta = 15^o`

The distance covered is
`d = 50 \ m`

The kinetic force is
`F_k = 255 \ N`

Generally the work done is mathematically represented as

`W = F * cos(\theta ) * d`

=>
`W = 2511 * cos( 15 ) * 50`

=>
`W = 121271.99 \ J`

Generally according to the work-energy theorem

`T_w = \Delta KE`

Here
`T_w` is total work done which is mathematically represented as

`T_w = F_k * d + F d cos(\theta)`

=>
`T_w = 255 * 50 + 121271.99`

=>
`T_w = 134021.99 \ J`

Also
`\Delta KE` is the change in kinetic energy which is mathematically represented as

`\Delta KE = (1)/(2) * m * [ v^2 - u^0 ]`

Here u is the initial velocity of the shed when at rest

`\Delta KE = (1)/(2) * 375 * [ v^2 - 0^0 ]`

=>
`\Delta KE = 187.5 v^2`

So

`134021.99 = 187.5 v^2`

=>
`v = √(714.783 )`

=>
`v = 26.74 \ m/s`