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dimulka [17.4K]
4 years ago
14

A 62 pound box is on an incline. Determine the minimum force P that would result in the box starting to slide up the incline. (i

nclude units with answer)
Physics
1 answer:
Anon25 [30]4 years ago
8 0

Answer:

F > W * sin(α)

Explanation:

The force needed for the box to start sliding up depends on the incline (α).

The external forces acting on the box would be the weight, the normal reaction and the lifting force that is applied to make it slide up.

These forces can be decomposed on their normal and tangential (to the slide plane) components.

The weight will be split into

Wn = W * cos(α) (in normal direction)

Wt = W * sin(α) (in tangential direction)

The normal reaction will be alligned with the normal axis, and will be equal to -Wn

N = -W* cos(α) (in normal direction)

To mke the box slide up, a force must be applied, that is opposite to the tangential component of the weight and at least a little larger

F > |-W * sin(α)| (in tangential direction)

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A block of mass m is pushed up against a spring with spring constant k until the spring has been compressed a distance x from eq
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Answer:d

Explanation:

Spring is compressed to a distance of x from its equilibrium position

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i.e. Work done by block W=\frac{1}{2}kx^2

therefore spring will also done an equal opposite amount of work on the block in the absence of external force

Thus work done by spring on the block W=-\frac{1}{2}kx^2

Thus option d is correct

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A 800 J <br> B -800 J <br> C 400 J <br> D -400 J <br> ??
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In certain ranges of a piano keyboard, more than one string istuned to the same note to provide extra loudness. For example, the
steposvetlana [31]

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let the frequency is f'.

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A 14.3-g bullet is fired into a 5.21 kg block of wood. The block is attached to a spring that has a spring force constant of 450
Natasha2012 [34]

Answer:

The initial speed of the bullet is v_{o} = 889.199\,\frac{m}{s}.

Explanation:

The collision between bullet and block is inelastic and let suppose that motion occurs on a horizontal surface, so that changes in gravitational potential energy can be neglected. Initially, the intial speed of the bullet-block system can be determined with the help of the Work-Energy Theorem and the Principle of Energy Conservation:

K = U_{k} + W_{loss}

\frac{1}{2}\cdot (5.224\,kg)\cdot v^{2} = \frac{1}{2}\cdot \left(450\,\frac{N}{m}\right)\cdot (0.22\,m)^{2} + (0.35)\cdot (5.224\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (0.22\,m)

The initial speed of the bullet-block system is:

v \approx 2.383\,\frac{m}{s}

Now, the initial speed of the bullet is determined by applying the Principle of Momentum Conservation:

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