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8090 [49]
3 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]3 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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For a given initial velocity, how does the time td it takes to stop on dry snow differ from the time tw it takes to stop on wet
salantis [7]

let the friction coefficient on dry ice and wet ice is different and given as

\mu_d = on dry ice

\mu_w = on wet ice

now by friction force formula we can say

F_f = -\mu mg

so the deceleration due to friction force is given as

a = -\mu g

now the deceleration due to friction force on both surface is given as

a_1 = -\mu_d * g

a_2 = -\mu_w * g

now the time to stop the two blocks can be calculated by kinematics

v_f - v_i = a t

0 - v_0 = -\mu_d * g * t_d

t_d = \frac{v_0}{\mu_d* g}

also for other block on wet ice we can say similarly

t_w = \frac{v_0}{\mu_w*g}

now the ratio of two time on two surfaces is given as

\frac{t_d}{t_w} = \frac{\mu_w}{\mu_d}

so this is the ratio of time on two surfaces

7 0
3 years ago
The force is proportional to what measurement
liberstina [14]
The extension (in metres), if looking at Hooke's law
5 0
3 years ago
Water from a fire hose is directed horizontally against a wall at a rate of 69.4 kg/s and a speed of 19.6 m/s. Calculate the mag
Gnoma [55]

Answer:

Force, |F| = 1360.24 N

Explanation:

It is given that,

Water from a fire hose is directed horizontally against a wall at a rate of 69.4 kg/s, \dfrac{m}{t}=69.4\ kg/s

Initial speed of the water, u = 19.6 m/s

Finally water comes to stop, v = 0

To find,

The magnitude of the force exerted on the wall.

Solution,

Let F is the force exerted on the wall. The product of mass and acceleration is called the force exerted. Using second law of motion to find it as :

F=m\dfrac{v-u}{t}

F=\dfrac{-mu}{t}

F=-69.4\ kg/s\times 19.6\ m/s

|F| = 1360.24 N

So, the magnitude of the force exerted on the wall is 1360.24 N.

5 0
3 years ago
Question 9 of 34
Rufina [12.5K]

Answer:

Option B) Displacement

6 0
2 years ago
The towline exerts a force of p = 4 kn at the end of the 20-m-long crane boom. if u = 30, determine the placement x of the hook
Illusion [34]

Answer:

The answer is 80 kN . m (clockwise)

Explanation:

As,

M = P x L

Here, the towline exerts a force is P.

Substituting P for 4000N.

M = -4000N x 20m

   = -80000N.m

   = 80kN.m

Maximum moment about the point O is 80kN.m (Clockwise)

7 0
3 years ago
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