Question

is 63 2 kg es a plane smooth A block of mass moves smeters on horizontal surface along the direction of force UN raised to 3 metess height the net work done block will be a by​

Answer 1

Answer:

Given,

Work done =300J, mass=2kg,height=10m,g=10m/s

2

we know that the work done =fd

300=(mgsinθ+f)×

sinθ

h

Work-done against the friction:

f.

sinθ

h

=300−mgsinθ×

sinθ

h

⇒300−mgh=300−2×10×10=100J

solution

Answer 2

Answer:

the solution is 100J. GOOD LUCK