Question

consider a inclined plane of length 25 m at an angle 15°.an object of mass 6 kg moves with velocity 10 ms^-1 in the inclined plane find the work done(in KJ)​

Answer 1

in this answer the value is not seen so please you can put your value yourself

Explanation:

From FBD,

F=mgsinθ+f

or F=mg

x

h

+f as(sinθ=

x

h

)

or Fx=mgh+fx

or W

F

=mgh+W

f

as work done = force × distance.

150=5×10×2+W

f

or W

f

=150−100=

Answer 2

Explanation:

we get a horizontal components of

$60 \sin(15)$

before finding workdone we have to find force and acceleration

${v}^{2} = {u}^{2} + 2as \\ 0 = {10}^{2} + 2 \times a \times 25 \\ \\ a = - 2m {s}^{2}$

and therefore

$f = ma \\ 6 \times - 2 \\ 12n \: backwrd$

$w \: = f \times d \\ (12 + \ \: 60sin(15) ) \times 25 \\ 339.01j$