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A constant force F, applied at an angle of 20° above the horizontal, pulls a 200 kg block, over a distance of 3 m, on a rough, horizontal floor as shown in the diagram below - NSC Physical Sciences - Question 5 - 2016 - Paper 1
Question 5
A constant force F, applied at an angle of 20° above the horizontal, pulls a 200 kg block, over a distance of 3 m, on a rough, horizontal floor as shown in the diagr... show full transcript
Worked Solution & Example Answer:A constant force F, applied at an angle of 20° above the horizontal, pulls a 200 kg block, over a distance of 3 m, on a rough, horizontal floor as shown in the diagram below - NSC Physical Sciences - Question 5 - 2016 - Paper 1
Step 1
5.1 Give a reason why the coefficient of kinetic friction has no units.
Answer
The coefficient of kinetic friction is a ratio of two forces (the frictional force to the normal force). As such, its value is derived from the division of these two quantities, which results in a dimensionless quantity. Therefore, the coefficient of kinetic friction has no units.
Step 2
Step 3
5.3 Draw a free-body diagram indicating ALL the forces acting on the block while it is being pulled.
Answer
To draw the free-body diagram, depict the following forces:
- The pulling force, ( F ), acting at a 20° angle from the horizontal.
- The weight of the block, ( mg ), acting vertically downwards.
- The normal force, ( N ), acting vertically upwards, equal to the weight adjusted for the vertical component of the pulling force.
- The frictional force, ( F_f ), acting horizontally opposite to the direction of motion.
Step 4
5.4 Show that the work done by the kinetic frictional force (W_k) on the block can be written as W_k = \( \mu_k (mg - F \sin 20°) \cdot 3 \) J.
Answer
The work done by the kinetic frictional force is given by:
[ W_k = F_f \cdot d ] Where:
- ( F_f = \mu_k N )
- ( N = mg - F \sin 20° )
- And the distance ( d = 3 ) m.
Thus, substituting these values:
[ W_k = \mu_k (mg - F \sin 20°) \cdot 3 ]
Step 5
5.5 Calculate the magnitude of the force F that has to be applied so that the net work done by all forces on the block is zero.
Answer
To find the force ( F ) such that the net work done is zero, we set:
[ W_{net} = W_g + W_k + W_f = 0 ] Where:
- ( W_g = 0 ) since there's no displacement in the vertical direction.
- Substitute ( W_k ) calculated earlier:
[ 0 = (0) + \left(\mu_k (mg - F \sin 20°) \cdot 3\right) + \left( F \cos 20° \cdot 3 \right) ]
Solving for ( F ) gives:
[ F = 388.88 \text{ N} ] Thus, the required force to make the net work zero is approximately 389 N.