A block of mass 2kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F

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Solution

$F = m a$
Since F is constant, a is also constant.Therefore from Kinematics:
$S = u t + \frac{1}{2} a t^{2}$
$10 = 0 + \frac{1}{2} a \times 4$ (Since object is at rest initially)
$a = \frac{10}{2} = 5 m / s^{2}$
$F = 2 \times 5 = 10 N$

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A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F.

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A block of mass 2kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F

F=maSince F is constant, a is also constant.Therefore from Kinematics:S=ut+12at210=0+12a×4 (Since object is at rest initially)a=102=5m/s2F=2×5=10N

$F = m a$
Since F is constant, a is also constant.Therefore from Kinematics:
$S = u t + \frac{1}{2} a t^{2}$
$10 = 0 + \frac{1}{2} a \times 4$ (Since object is at rest initially)
$a = \frac{10}{2} = 5 m / s^{2}$
$F = 2 \times 5 = 10 N$