A solid cylinder is lifted out of oil by a wire attached to a motor. Fig. 5.1 shows two different positions X and Y of the cylinder during the lifting process. [Figure 5.1] The motor is fixed to an overhead beam. The cylinder has cross-sectional area 0.018m², length 1.2m and weight 560 N. The density of the oil is 940kgm⁻³. Throughout the lifting process, the cylinder moves vertically upwards with a constant velocity of 0.020 m s⁻¹. The viscous force of the oil acting on the cylinder is negligible.
- (a)Calculate the density of the cylinder.[2]
- (b)For the cylinder at position X, show that the upthrust due to the oil is 200 N.[2]
- (c)(i)Calculate, for the moving cylinder at position X, the tension in the wire,[1]
- (c)(ii)the power output of the motor.[2]
- (d)(i)The cylinder is raised with constant velocity from position X to position Y. State and explain the variation, if any, of the power output of the motor as the cylinder is raised. Numerical values are not required.[3]
- (d)(ii)The rate of energy output of the motor is less than the rate of increase of gravitational potential energy of the cylinder. Without calculation, explain this difference.[1]
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Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [C1] ρ = m/V
- [A1] = (560/9.81)/(1.2 × 0.018) = 2600 kgm⁻³
- [C1] (A)ρg = 940 × 9.81 × 1.2
- [A1] (upthrust =) 940 × 9.81 × 1.2 × 0.018 = 200 N
- [A1] tension = 560 – 200 = 360 N
- [C1] P = Fv
- [A1] = 360 × 0.020 = 7.2 W
- [B1] upthrust decreases
- [M1] tension (in wire) increases
- [A1] power (output of motor) increases
- [B1] there is work done (on the cylinder) by the upthrust OR GPE of oil decreases (as it fills the space left by cylinder and so total energy is conserved)
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