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Investigating Forces & Motion
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Investigating Forces and Motion (1998)(Granada Learning).iso
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topic6
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example.dat
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1998-02-16
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83 lines
[general]
[page1]
type:0
caption:\
<img src="6ex1" align=center><p>\
A truck with a mass of 5.0 kg is accelerated by an average force of 20 \
N along a 3.0m track. Calculate the truck's kinetic energy and its \
speed at the end of the track.<p>\
work done = <i>F</i> x <i>d</i><p>\
= 20 x 3.0<p>\
= 60 J<p>\
Asuming negligible friction, the work done is equal to the kinetic \
energy gained, so:<p>\
<i>E<sub>k</sub></i> = 60 J<p>\
If kinetic energy = <i>½mv<sup>2</sup></i>, then:<p>\
<i>v</i> = <img src="6sqrt">(2 <i>E<sub>k</sub> / m</i>).<p>\
= <img src="6sqrt">(2 x 60 / 5)<p>\
= 4.9 m/s<p>
[page2]
type:0
caption:\
<img src="6ex2" align=center><p>\
A steel ball (m = 0.5 kg) is released from rest on a curved \
semicircular chute. It rolls to and fro, up and down. The vertical \
distance from the top of the chute to the base is 1.0 m.<p>\
What is the ball's potential energy at the top of the chute?<p>\
Using <i>E<sub>g</sub></i> = <i>mgh</i> and taking <i>g</i> = 10 \
m/s<sup>2</sup>,<p>\
<i>E<sub>g</sub></i> = 0.5 x 10 x 1.0 = 5.0 J<p>
[page3]
type:0
caption:\
<img src="6ex2" align=center><p>\
A steel ball (m = 0.5 kg) is released from rest on a curved \
semicircular chute. It rolls to and fro, up and down. The vertical \
distance from the top of the chute to the base is 1.0 m.<p>\
Assuming friction is negligible, use the priciple of conservation of \
energy to calculate the ball's kinetic energy at the bottom of the \
chute.<p>\
In the absence of friction, the total energy of the ball is constant. \
Initially, the ball has 5.0 J of potential energy (see example 2). As \
it falls and speeds up, its potential energy is converted into kinetic \
energy. At the bottom of the chute, all the potential energy has been \
converted into kinetic energy, so:<p>\
<i>E<sub>k</sub></i> = 5.0 J<p>
[page4]
type:0
caption:\
<img src="6ex4" align=center><p>\
A steel ball (m = 0.5 kg) is released from rest on a curved \
semicircular chute. It rolls to and fro, up and down. The vertical \
distance from the top of the chute to the base is 1.0 m.<p>\
What are the ball's potential and kinetic energies when it is at point \
X halfway down the chute?<p>\
At the top of the chute the ball's potential energy is 5.0 J (see \
example 2). Halfway down the chute, h = 0.5 m, and the ball's \
potential energy is given by:<p>\
<i>E<sub>g</sub> = mgh</i> = 0.5 x 10 x 0.5 = 2.5 J<p>\
Now, since energy is conserved,<p>\
potential energy loss = kinetic energy gain.<p>\
Therefore:<p>\
<i>E<sub>k</sub></i> = 5.0 - 2.5 = 2.5 J<p>
[page5]
type:0
caption:\
A weightlifter lifts a 100 kg mass through a height of 2.0 m in 2.0 s. \
What is the average power of his lift? (Assume <i>g</i> = 10 \
m/s<sup>2</sup>).<p>\
By lifting weights, the lifter increases their gravitational potential \
energy:<p>\
increase in potential energy = <i>mgh</i><p>\
= 100 x 10 x 2.0<p>\
= 2 000 J<p>\
power = energy transferred / time<p>\
= 2 000 / 2.0<p>\
= 1 000 W (or 1 kW)<p>