Whilst being raised, the load accelerates uniformly upwards. The weight of the cable is negligible compared to the weight of the load.

(i) Draw a labelled free-body force diagram of the forces acting on the accelerating load. The dot below represents the load.

(ii) The load has a mass of 350 kg and it takes 6.5 s to raise it from rest through a height of 8.0 m.

Determine the tension in the cable as the load is being raised.

The electric motor can be adjusted such that, after an initial acceleration, the load moves at constant speed. The motor is connected to a 450 V supply and with the load moving at constant speed, it takes the motor 15 s to raise the load through 7.0 m.

(i) Calculate the power delivered to the load by the motor.

(ii) The current in the motor is 30 A. Estimate the efficiency of the motor.

(i) upward arrow labelled T/tension/force in cable and downward arrow labelled W/mg/weight/gravity force;{ (both needed)
tension arrow length >weight length;

(ii) \(a = \frac{{2s}}{{{t^2}}}\);
\(a = \left( {\frac{{2 \times 8.0}}{{{{6.5}^2}}} = } \right)0.38\left( {{\rm{m}}{{\rm{s}}^{ - 2}}} \right)\);
T=ma+mg or T=350(0.38+9.8);
3.6 kN;
Allow g=10 N kg^-1 (same answer to 2 sf).

(i) change in gpe=350×9.81×7.0(=24kJ);
power \(\left( { = \frac{{24 \times {{10}^3}}}{{15}}} \right) = 1.6{\rm{kw}}\);
Allow g=10Nkg^-1.

(ii) power input to motor=13.5 (kW);
efficiency=\(\left( {\frac{{1.6}}{{13.5}} = } \right)0.12\) or 12%;