Q.A body of mass 5 kg, initially at rest, is moved by a horzantal force of 2 N on a smooth horizantal surface. Find the work done by the force in 10 s
A.10 J
B. 20J
C.30J
D.40J
E.25J
Answers
Answered by
4
=>F = ma
=>2N = 5× a
So , a = 2/5=0.4 m /s²
Now, u =0
a = 2/5=0.4 m /s²
And t =10 s
=>S =ut +1/2at²
=>S= 0×10 +1/2×2/5×(10)²
=>S= 1/5×10×10 = 20m.
=>Work done = f × s
=>Work done = 2×20 =40J
=>2N = 5× a
So , a = 2/5=0.4 m /s²
Now, u =0
a = 2/5=0.4 m /s²
And t =10 s
=>S =ut +1/2at²
=>S= 0×10 +1/2×2/5×(10)²
=>S= 1/5×10×10 = 20m.
=>Work done = f × s
=>Work done = 2×20 =40J
Answered by
0
40 joules as a= f/m a is 0.4 m/s2 v= u+at v= 4m/s work dine= 1/2 mv square ans is 40 joules
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