Question

A block of mass 2.50kg is pushed 2.20 m along a frictionless horizontal table by a constant 16.0 N force directed 25.0° below the horizontal determine the work done on the block by the applied of B. The normal force exerted by the table
C. Gravitational force D. The total work done

Answer 1

Explanation:

A. work done = f.s

w = 16×2.20

w = 35.2 Nm

B. mass×g

C. 2.50×9.8 = 24.5

D. w = f.s

w = 25.0 ×2.20

w = 55

Answer 2

Work done.

Explanation:

Given that,

Mass of the block, m = 22.5 kg

Distance, d = 2.2 m

Force acting on the block, F = 16 N

Angle is, $\theta=25^{\circ}$

(A) Let W is the work done on the block by the applied force. It is given by :

$W=Fd\cos\theta\\\\W=16\times 2.2\times \cos25\\\\W=31.9\ J$

(B) As the normal force is perpendicular to the displacement of the block. So, W = 0

(C) Gravity force is also perpendicular to the displacement. So,we have work done is equal to 0.

(d) Total work done is 31.9 J + 0 + 0 = 31.9 Joules

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Work done

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