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Aleks [24]
3 years ago
5

A grocery shopper tosses a(n) 8.5 kg bag of rice into a stationary 17.7 kg grocery cart. The bag hits the cart with a horizontal

speed of 6.9 m/s toward the front of the cart. What is the final speed of the cart and bag? Answer in units of m/s.
Physics
1 answer:
timama [110]3 years ago
4 0

Answer:

2.2386 m/s

Explanation:

I think you would use the momentum equation.

m1v1=m2v2

m1= mass of the bag (8.5kg)

v1= speed of bag (6.9 m/s)

m2= mass of bag and cart (26.2 kg)

v2= speed of cart and bag

Plug these numbers all into the equation and you solve for v2 which ends up at 2.2386 m/s

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A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput
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There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span>E=U+K=  \frac{1}{2}kx^2 + \frac{1}{2} mv^2
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m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span>E=K_{max} =  \frac{1}{2}mv_{max}^2
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
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<span>
Since the mechanical energy must be conserved, we can write
</span>\frac{1}{2}mv_{max}^2 =  \frac{1}{2}kA^2
<span>from which we find the maximum speed
</span>v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }=  1.2 m/s
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part. 
The total mechanical energy is:
</span>E=K_{max}=  \frac{1}{2}mv_{max}^2= 0.36 J
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span>U= \frac{1}{2}kx^2 =  \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J
<span>And since 
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<span>we find the kinetic energy when the glider is at this position:
</span>K=E-U=0.36 J - 0.05 J = 0.31 J
<span>And then we can find the corresponding velocity:
</span>K= \frac{1}{2}mv^2
v=  \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
a_{max} = \omega^2 A
where \omega= \sqrt{ \frac{k}{m} } is the angular frequency, and A is the amplitude.
The angular frequency is:
\omega =  \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s
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a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2

d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
</span>a(t)=\omega^2 x(t)
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</span>a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span>E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J
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