A force F -5 -6 -4 k- acting on a body- produces a displacement s -6 i -5 k- Work done by the force isA- 18 unitsB- 12 unitsC- 15 unitsD- 10 units

A force $\to F = 4^i + 6^j + 3^k$ acting on a particle produces a displacement of $\to S = 2^i + 3^j + 5^k$ where F is expressed in Newtonn and S in the metre. Find the work done by the force.

Q. The work done by the force

(4, - 2, 3)

to move a particle along a straight line from the point

A (3, 3, - 1)

B (6, 4, - 2)

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A force →F=5^i+6^j−4^k acting on a body, produces a displacement →s=6→i+5→k. Work done by the force is

The correct option is D 10 units W=→F.→s=(5^i+6^j−4^k).(6^i+5^k)=30−20=10 units

A force $\to F = 5^i + 6^j - 4^k$ acting on a body, produces a displacement $\to s = 6 \to i + 5 \to k$ . Work done by the force is

The correct option is D
10 units

$W = \to F . \to s = (5^i + 6^j - 4^k) . (6^i + 5^k) = 30 - 20 = 10 u n i t s$