A moving 2.8 kg block collides with a horizontal spring whose spring constant is 215 N/m (see figure). The block compresses the spring a maximum distance of 11.5 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.48
How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?
I found the work done by the spring -1.42J
I get the friction work to be
(friction force)(distance moved) =
(0.48)(2.8)(9.8)(0.115) = 1.52 J
You don't need the spring constant to get the answer.
the next part is
What is the speed of the block when it hits the spring?
i did
1.42=.5mv^2(final)-.5mv^2(initail)
The final velocity woulf be zero
so
1.42=.5(2.8)v^2
v=1.01
im not sure what i did wrong
Initial KE = (stored spring PE when stopped) + friction work
(1/2)MV^2 = (1/2)kX^2 + 1.52 J
Now recalculate V. Include the units. Numbers by themselves, without units, are useless.
thanks
Why is friction work positive????
To find the work done by the force of friction, we can begin by finding the work done by the spring.
The work done by the spring is given by the formula:
\(W_{\text{spring}} = -\frac{1}{2} k x^2\)
where \(k\) is the spring constant and \(x\) is the compression distance.
In this case, the spring constant \(k\) is 215 N/m and the compression distance \(x\) is 11.5 cm. However, we need to convert the compression distance to meters:
\(x = 11.5 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 0.115 \, \text{m}\)
Now we can plug in the values:
\(W_{\text{spring}} = -\frac{1}{2} \times 215 \, \text{N/m} \times (0.115 \, \text{m})^2\)
\(W_{\text{spring}} = -1.42 \, \text{J}\)
So, the work done by the spring is -1.42 J.
Now, to find the work done by the force of friction, we can use the formula:
\(W_{\text{friction}} = f_{\text{friction}} \times d\)
where \(f_{\text{friction}}\) is the force of friction and \(d\) is the distance over which the force is applied.
We can find the force of friction by using the equation:
\(f_{\text{friction}} = \mu \times f_{\text{normal}}\)
where \(\mu\) is the coefficient of kinetic friction and \(f_{\text{normal}}\) is the normal force.
The weight of the block is given by:
\(f_{\text{normal}} = m \times g\)
where \(m\) is the mass of the block and \(g\) is the acceleration due to gravity.
In this case, the mass \(m\) is 2.8 kg and \(g\) is approximately 9.8 m/s². Plugging in the values, we can find:
\(f_{\text{normal}} = 2.8 \, \text{kg} \times 9.8 \, \text{m/s}^2\)
\(f_{\text{normal}} = 27.44 \, \text{N}\)
Now, we can calculate the force of friction using the coefficient of kinetic friction:
\(f_{\text{friction}} = 0.48 \times 27.44 \, \text{N}\)
\(f_{\text{friction}} = 13.1712 \, \text{N}\)
Finally, we can find the work done by the force of friction by multiplying the force of friction by the distance over which it is applied. In this case, the compression distance of the spring is also the distance over which the force of friction is applied, so:
\(W_{\text{friction}} = f_{\text{friction}} \times x\)
\(W_{\text{friction}} = 13.1712 \, \text{N} \times 0.115 \, \text{m}\)
\(W_{\text{friction}} \approx 1.514 \, \text{J}\)
Therefore, the amount of mechanical energy dissipated by the force of friction while the block is being brought to rest by the spring is approximately 1.514 J.