Question024 (part 1 of 3 ) 10.0 polnts A 4.0 kg block is pushed 2.0 m at a constant velocity up a vertical wall by a constant force applied at an angle of $26.0^{\circ}$ with the horizontal, as shown in the figure.
The acceleration of gravity is $9.81 \mathrm{~m} / \mathrm{s}^{2}$.
Drawing not to scale. If the coefficient of kinetic friction between the block and the wall is 0.20 , find a) the work done by the force on the block. Answer in units of J.
025 (part 2 of 3 ) 10.0 points b) the work done by gravity on the block. Answer in units of J.
026 (part 3 of 3 ) 10.0 points c) the magnitude of the normal force between the block and the wall.
Answer in units of N .

Studdy Solution
Calculate the normal force between the block and the wall:
3.1. Use the horizontal component of the applied force: $$N = F \cos(26^\circ)$$
Now, let's solve each part:
STEP_1: 1.1. Assume $F$ is the magnitude of the applied force. 1.2. Since the block moves at constant velocity, the net force is zero: $$F \sin(26^\circ) = mg + \mu_k F \cos(26^\circ)$$ $$F \sin(26^\circ) = 4.0 \times 9.81 + 0.20 \times F \cos(26^\circ)$$
1.3. Solve for $F$: $$F (\sin(26^\circ) - 0.20 \cos(26^\circ)) = 4.0 \times 9.81$$ $$F = \frac{4.0 \times 9.81}{\sin(26^\circ) - 0.20 \cos(26^\circ)}$$
1.4. Calculate the work done by the force: $$W_{\text{force}} = F \cdot 2.0$$
STEP_2: 2.1. Calculate the work done by gravity: $$W_{\text{gravity}} = -4.0 \times 9.81 \times 2.0$$ $$W_{\text{gravity}} = -78.48 \, \text{J}$$
STEP_3: 3.1. Calculate the normal force: $$N = F \cos(26^\circ)$$
The solutions are: a) Work done by the force: $W_{\text{force}}$ b) Work done by gravity: $-78.48 \, \text{J}$ c) Normal force: $N$