Fig. 9.1 shows a block of mass 5.0 kg resting on a rough surface. A horizontal rope applies a pulling force. The block does not move. (a) Calculate the weig...
Question 1 Report
Fig. 9.1 shows a block of mass 5.0 kg resting on a rough surface. A horizontal rope applies a pulling force. The block does not move.
(a) Calculate the weight of the block. Use g = 10 N/kg. [1]
(b) The block is pulled with a horizontal force of 12 N but does not move. State the size of the friction force. [1]
(c) State the resultant force on the block. [1]
(d) The pulling force is increased to 20 N. The block begins to accelerate at 1.6 m/s². Calculate the friction force now acting. [2]
(e) Draw a labelled force diagram showing all forces acting on the block in part (d). [2]
Answer Details
(a) Using \( W = mg \):
\[ W = 5.0 \times 10 = 50 \text{ N} \]
[1]
(b) The block does not move, so the resultant horizontal force must be zero. The friction force must therefore equal the pulling force:
\[ F_{\text{friction}} = 12 \text{ N (to the left)} \]
[1]
Friction adjusts up to a maximum value to match the applied force and prevent motion.
(c) The resultant force on the block is zero. [1]
The block is stationary, so all forces are balanced: horizontally, pull = friction; vertically, weight = normal reaction.
(d) First, find the resultant force using \( F = ma \):
\[ F_{\text{resultant}} = ma = 5.0 \times 1.6 = 8.0 \text{ N (to the right)} \]
[1]
The resultant is the difference between the pull and friction:
\[ F_{\text{resultant}} = F_{\text{pull}} - F_{\text{friction}} \]
\[ 8.0 = 20 - F_{\text{friction}} \]
\[ F_{\text{friction}} = 20 - 8.0 = 12 \text{ N} \]
[1]
(e) Force diagram for part (d): [2]
Four forces act on the block: weight (50 N downward), normal reaction (50 N upward), pull (20 N to the right), and friction (12 N to the left). The vertical forces balance. The horizontal forces do not balance, giving a resultant of 8.0 N to the right, which causes the acceleration.
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