Physics, asked by vijaya2002, 1 year ago

work done in moving a 10kg mass from bottom to top of smooth inclined surface of angle 30 through a distance of 5m is

Answers

Answered by Anonymous

2

work done=F*d=mg * sin@ *d=10*10*0.5*5=250J

Answered by kinjalsheth79

1

Answer:

250 J

Explanation:

w= Fdcos@

=mgsin@*dcos@

=mgsin30°*dcos0°

=10*10*1/2*5*1

W=250J

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