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cupoosta [38]
3 years ago
13

A toy dart gun has a spring with k= 128 N/m. How much force does it take to pull the spring back 0.0500 m? (Unit = N)​

Physics
2 answers:
poizon [28]3 years ago
7 0

Answer:

6.40

Explanation:

Acellus Proof

yKpoI14uk [10]3 years ago
4 0

Answer:

F = 0.64 N

Explanation:

We are given;

Spring constant constant; k = 1.28 N/m

Distance; x = 0.5 m

From Hooke's law, we know that F = kx.

Thus;

F = 1.28 × 0.5

F = 0.64 N

Thus, force it takes to pull the spring back = 0.64 N

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Explanation:

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Two weights are connected by a very light cord that passes over an 80.0Nfrictionless pulley of radius 0.300m. The pulley is a so
Citrus2011 [14]

Answer:

The force does the ceiling exert on the hook is 269.59 N

Explanation:

Applying the second Newton law:

F = m*a

From the attached diagram, the net force in object 1 is:

m_{1} a=T_{1} -W_{1}

In object 2:

m_{2} a=W_{2} -T_{2}

Adding the two equations:

m_{2} a+m_{1} a=T_{1} -W_{1} +W_{2} -T_{2} \\m_{1} =\frac{W_{1} }{g} \\m_{2} =\frac{W_{2} }{g} \\Replacing\\T_{2}-T_{1}=W_{2}   -W_{1} -(\frac{W_{1} }{g} +\frac{W_{2} }{g} )a  (eq. 1)

The torque:

\tau =I\alpha

Where

I = moment of inertia

α = angular acceleration

If the linear acceleration is

a=r\alpha \\\alpha =\frac{a}{r} \\I=\frac{1}{2} mr^{2} \\\tau =\frac{mra}{2}

Torque due the tension is equal:

\tau =r(T_{2} -T_{1} )

Substituting torque, mass, in equation 1, the expression respect the acceleration is:

a=\frac{g*(W_{2}-W_{1})}{W_{1}+W_{2} +\frac{W}{2} }

Where

W₁ = 75 N

W₂ = 125 N

W = 80 N

a=\frac{9.8*(125-75)}{75+125+\frac{80}{2} } =2.04m/s^{2}

The net force is:

F_{n} =F-W-T_{1} -T_{2}\\0=F-W-W_{1} (\frac{a}{g} +1)-W_{2} (1-\frac{a}{g})\\F=W+W_{1} +W_{2} +\frac{a}{g} (W_{1} -W_{2} )\\F=80+75+125+\frac{2.04}{9.8} (75-125)\\F=269.59N

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