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8090 [49]
2 years ago
12

A 56 kg object is attached to a rope, which can be used to move the load vertically.

Physics
1 answer:
Sergio039 [100]2 years ago
8 0

Answer:

a) T=549.36N Upwards

b) T=448.56N Upwards

c) T=650.16N Upwards

Explanation:

The very first thing we can do to solve this problem is to draw a free body diagram we can use to analyze the situation (see attached picture).

On the diagram we can see there are only two forces acting on the object: the tension of the rope and the weight of the object itself.

a)

Since the object is moving at a constant speed, this means that its acceleration will be zero. So we can do a sum of forces like this:

\sum F=0

T-W=0

T=W

T=mg

T=(56kg)(9.81m/s^{2})

T=549.36N upwards

b)

For part b, since the object is accelerating downwards, we wil say that it's acceleration is negative, so a=-1.8m/s^{2}

so we can do a sum of forces again:

\sum F=ma

so

T-W=ma

T=ma +W

T=ma+mg

T=m(a+g)

and now we substitute:

T=(56kg)(-1.8 m/s^{2}+9.81 m/s^{2})

which yields:

T=448.56N upwards (in this particular case, the tension always goes upwards)

c)

Since the object is moving upwards, we can say that its acceleration will be positive, so a =1.8m/s^{2}

we can take the solved equation we got on the previous part of the problem, so we get:

T=m(a+g)

T =(56kg)(1.8 m/s^{2}+9.81 m/s^{2})

which yields:

T=650.16N upwards

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It measure the effort it takes for an object to move

Explanation:

Work done is the force applied to move a body through a particular distance.

It can also to be said to be a measure of the effort it takes for an object to move.

The unit of work done is in Joules.

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Answer:

An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. ... A net force = unbalanced force. If however, the forces are balanced (in equilibrium) and there is no net force, the object will not accelerate and the velocity will remain constant.

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To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

where

h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

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The cargo reaches the ground when

y(t) = 0

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

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b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

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t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

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