- #1
Peter G.
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The angle of the slope is 4.8 with the horizontal and the weight of the lorry is 2.4 x 105. A resistive force of 1.2 x 104 down the slope acts on the lorry as it travels up the slope at a constant speed of 15 m/s
(a) Show that the component of the weight of the lorry down the slope is 2.0 x 104:
(b) Calculate the rate at which the truck does work against the resistive forces:
(c) Calculate the power developed by the lorry as it travels up the slope:
(d) Calculate the rate at of gain of potential energy of the lorry:
(e) State and explain how the braking distance of the lorry up the slope compares with that on a horizontal road at the same speed:
The way question c and d are written confuse me a bit. These were my attempts but I am unsure whether they are right or not:
(a) I always have difficulty determining the angle when dividing the components of weight, but: I figured: sin 4.8 x 2.4 x 105 = 2.0 x 104
(b) P = Force x Velocity, P = 1.2 x 104 x 15 = 180,000 J/s
(c) Constant force, hence, force back - resistive forces and the component of weight parallel to the slope - equal the force of the engine, thus: P = F x v, P = 252,000 x 15 = 3,780,000 W
(d) m x g x h, thus: 2.0 x 105 x (sin 4.8 x 15) = 301,240.2 J/s
(e) I think the braking distance will be reduced. On the horizontal road, the resistive forces are equal, provided, as stated in the question, the speed is the same, but, on the slope, there is also the component of the weight parallel to the truck pushing it back which will reduce the braking distance
Thanks in advance,
Peter G.
(a) Show that the component of the weight of the lorry down the slope is 2.0 x 104:
(b) Calculate the rate at which the truck does work against the resistive forces:
(c) Calculate the power developed by the lorry as it travels up the slope:
(d) Calculate the rate at of gain of potential energy of the lorry:
(e) State and explain how the braking distance of the lorry up the slope compares with that on a horizontal road at the same speed:
The way question c and d are written confuse me a bit. These were my attempts but I am unsure whether they are right or not:
(a) I always have difficulty determining the angle when dividing the components of weight, but: I figured: sin 4.8 x 2.4 x 105 = 2.0 x 104
(b) P = Force x Velocity, P = 1.2 x 104 x 15 = 180,000 J/s
(c) Constant force, hence, force back - resistive forces and the component of weight parallel to the slope - equal the force of the engine, thus: P = F x v, P = 252,000 x 15 = 3,780,000 W
(d) m x g x h, thus: 2.0 x 105 x (sin 4.8 x 15) = 301,240.2 J/s
(e) I think the braking distance will be reduced. On the horizontal road, the resistive forces are equal, provided, as stated in the question, the speed is the same, but, on the slope, there is also the component of the weight parallel to the truck pushing it back which will reduce the braking distance
Thanks in advance,
Peter G.
