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andrew-mc [135]
2 years ago
6

A weightlifter takes 0.5s to raise a barbell with 2 free-weights from the ground to a height of 0.9m. If each free-weight has a

mass of 30kg and the barbell has a mass of 20kg, calculate how much power the weightlifter produces when lifting the barbell?​
Physics
1 answer:
Anna [14]2 years ago
5 0

Answer:

P = 1412.82 [Watt]

Explanation:

First we must calculate the total mass that makes up the barbell and the two free weights at each end of the barbell.

m=30+30+20\\m= 80 [kg]

Weight is now defined as the product of mass by gravitational acceleration.

w=m*g

where:

m = mass = 80 [kg]

g = gravity acceleration = 9.81 [m/s²]

w=80*9.81\\w=784.8[N]

Now using the product of the weight by the distance traveled we can calculate the work.

W=w*d

where:

W = work [J]

w = weight = 784.8 [N]

d = distance = 0.9 [m]

W=784.8*0.9\\W=706.41[J]

And power is defined as the relationship of work at a certain time. The potency is expressed by means of the following formula.

P=W/t\\P=706.41/0.5\\P=1412.82[W]

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A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The
lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

E_{i} = E_{f}

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}                              

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s      

 

3) When the object is at the equilibrium position, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s

I hope it helps you!                                                                                        

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