A van of total mass 1800 kg travels at a uniform speed of 50 km.hr-1 along a road with a 1:50 gradient, against constant frictional forces totalling 450 N. How much work has been done by the vehicle in travelling 1 km? (g = 9.80 m.s-2.)
A van of total mass 1800 kg travels at a uniform speed of 50 km.hr-1 along a road with a 1:50 gradient, against constant frictional forces totalling 450 N. How much work has been done by the vehicle in travelling 1 km? (g = 9.80 m.s-2.) |
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The van is performing work against two forces:
Against the forces of friction: 450 (N) x 1000 (m) = 450 kJ.
Against gravity: For every 50 metres travelled along the road, the van climbs 1 metre. So, in travelling 1 km (1000 m), the van will have climbed a vertical distance of 20 m. The work done will therefore be: mgh = 1800 (kg) x 9.80 (m.s-2) x 20 (m) = 352800 J = 353 kJ.
The total work done will therefore be: 353 + 450 = 803 kJ
Note: the word "gradient" can be misleading here. Normally it means the tangent of the angle that a sloping line makes with the horizontal. In this problem, we have taken the value 1/50 to be the sine of the angle, that is, the distance is measured along the road. For small angles, it makes no significant difference, as sinα ≅ tanα for small values of α (less than 5º)
