An inclined plane is 5 metres long and 2 metres high. A force of 50 N is required to push a box up along the length of the slope.
Assuming no friction and 100% efficiency, what is the mass of the box to the nearest kilogram?
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An inclined plane is 5 metres long and 2 metres high. A force of 50 N is required to push a box up along the length of the slope.
Assuming no friction and 100% efficiency, what is the mass of the box to the nearest kilogram?
\(B\)
\(\text{Energy moving box}\ = 50 \times 5 = 250\ \text{J}\)
\(\text{Energy used = increase in potential energy of box (100% efficiency)}\)
| \(m \times g \times h\) | \(=250\) | |
| \(m\) | \(=\dfrac{250}{10 \times 2}=12.5\ \text{kg}\) |
\(\Rightarrow B\)
A box with weight `W` and subject to a friction force `F` is being pulled up an inclined plane by a force `P`.
The force `R` is the resultant of which two forces?
`D`
`=>D`
A child and sled with a combined mass of 23 kg are being pulled along a horizontal snow-covered surface using a rope.
The coefficient of static friction between the sled and the snow is 0.14.
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The diagram shows the forces acting on a block on an inclined plane. The block is held in a stationary position by force `P`.
What is the correct equation for the normal force, `N` ?
`A`
The diagram shows a box which is at rest on an inclined ramp. The length of the ramp is 3 m and one end is raised `h` metres off the ground.
The coefficient of friction between the ramp and the box is 0.5.
To what height `(h)` must the ramp be raised for the box to just begin to slide? (3 marks)
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`1.34\ text{m}`
| `mu` | `= 0.5 = tan\ theta` | |
| `theta` | `= tan^(-1)(0.5)= 26.57°` |
| `sin(26.57°)` | `= h/3` | |
| `:.h` | ` = 3 xx sin(26.57°)= 1.34\ text{m}` |
A skateboard rider is at rest at the top of a slope, indicated by point `A` in the diagram.
The combined mass of the rider and skateboard is 80 kg.
Calculate the speed of the rider, in metres per second, at point `B` if the average work done by friction against the skateboard is 55 N. (4 marks)
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`text{Speed} = 14.57\ text{m/s}`
`m= 80\ text{kg}, \ f= 55\ text{N}`
`PE= mgh= 80xx10xx27.8= 22\ 240\ text{J}`
`text{Losses to friction (W):}`
`W= fs= 55xx250= 13\ 750\ text{J}`
| `text{Energy used}` | `= PE−text{Frictional losses}` | |
| `= 22\ 240−13\ 750` | ||
| `= 8490\ text{J}` |
`text{Calculate speed at point B:}`
| `1/2×m×v^2` | `= 8490` | |
| `0.5×80×v^2` | `= 8490` | |
| `v^2` | `= 212.25` | |
| `:.v` | `= 14.57\ text{m/s}` |