Question

a block of mass m = 2.50 kg is pushed d = 2.40 m along a frictionless horizontal table by a constant applied force of magnitude f = 15.0 n directed at an angle θ = 25.0° below the horizontal as shown in the figure. (a) determine the work done by the applied force. (b) determine the work done by the normal force exerted by the table. (c) determine the work done by the force of gravity. (d) determine the work done by the net force on the block.

Explanation:

Step1: Recall work - formula

The work - done formula is $W = Fd\cos\theta$, where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and the displacement.

Step2: Calculate work done by the applied force (a)

Given $F = 15.0\ N$, $d=2.40\ m$, and $\theta = 25^{\circ}$.
$W_a=Fd\cos\theta=15.0\times2.40\times\cos(25^{\circ})$
$W_a = 15.0\times2.40\times0.9063\approx32.6\ J$

Step3: Analyze the normal force (b)

The normal force $\vec{N}$ is perpendicular to the displacement of the block (which is horizontal). Since $\theta = 90^{\circ}$ for the normal force and displacement, using $W = Fd\cos\theta$, and $\cos(90^{\circ}) = 0$.
$W_b = N\times d\times\cos(90^{\circ})=0\ J$

Step4: Analyze the force of gravity (c)

The force of gravity $\vec{F}_g$ acts vertically downwards and the displacement is horizontal. So, $\theta = 90^{\circ}$ for the force of gravity and displacement. Using $W = Fd\cos\theta$, and $\cos(90^{\circ}) = 0$.
$W_c = F_g\times d\times\cos(90^{\circ})=0\ J$

Step5: Calculate the net force (d)

Since there is no friction and the only non - zero horizontal component of force is the horizontal component of the applied force, the net force in the direction of displacement is $F_{net,x}=F\cos\theta$.
$W_d = F_{net,x}d=Fd\cos\theta$.
$W_d=15.0\times2.40\times\cos(25^{\circ})\approx32.6\ J$

Answer:

(a) $32.6\ J$
(b) $0\ J$
(c) $0\ J$
(d) $32.6\ J$