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A 1.8kg block is moved at constant speed...

A `1.8kg` block is moved at constant speed over a surface for which coefficient of friction `mu=1/4` it is pulled by a force F acting at `45^@` with horizontal as shown in Fig. The block is displaced by 2 m. Find the work done on the block by (a) the force F (b) friction (c) gravity.
.

A

`36/5 J , -36/5 J, 0`

B

`5/36 J , -36/5 J, 0`

C

`36/5 J , 36/5 J, 0`

D

`-36/5 J , -36/5 J, 0`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`N=mg-F sin 45^@=18-F/(sqrt2)`
Moving with constant speed means net force=0.
`F cos 45^@-muN=1/4(18-F/(sqrt2))`
`:. (4F)/(sqrt2)=18-(F)/(sqrt2)`
`:. F=(18sqrt2)/(5)N`
(a) `W_(F)=FS cos45^@`
`=((18sqrt2)/5)(2)(1/(sqrt2))=7.2J`
(b) `W_(f)=(muN)(S) cos 180^@`
`=(1/4)(18-(F)/(sqrt2))(2) (-2)`
(c) `W_(mg)=(mg)(s) cos 90^@ =0`.
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