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matrenka [14]
3 years ago
12

A 200g(m2) mass is pulling the cart (m1) across the table as shown below. If it covers 1.12m in 4.8 seconds starting from rest,

what is the mass of the cart, assuming no friction?

Physics
1 answer:
Vsevolod [243]3 years ago
4 0

Answer: 20.21 kg

Explanation:

The mass hanging from the pulley is pulling the cart. The force by which cart is being pulled is equal to weight of the hanging mass.

F=m_2g

⇒F=0.200 kg ×9.8 m/s²=1.96 N

Since there is no other force acting on the cart and there is no friction, so this force will pull the cart.

The Cart covers 1.12 m distance in 4.8 s.

From the equation of motion,

s = u t + 0.5 × a t²

We will find the acceleration of the cart,

Distance covered, s = 1.12 m

Initial velocity, u = 0 ( staring from rest)

Time taken, t = 4.8 s

⇒1.12 m = 0 + 0.5 × a × (4.8 s)²

⇒a = 0.097 m/s²

Now the force which causes this acceleration is:

F = ma

where m is the mass of the cart and a is the acceleration of the cart

⇒1.96 N = m×0.097 m/s²

⇒m = 20.21 kg

Hence, the mass of the cart is 20.21 kg.


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To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

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The distance in this case is a composition between number of steps and the height. Then,

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On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

V_f^2-V_i^2 = 2a\Delta X

Where,

V_f = Final velocity

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PART A) For the particular case of work we know then that,

W = F*d

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W = 50*9.8*(0.3*30)

W = 4.41kJ

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V_f^2-V_i^2 = 2a\Delta X

Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

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Replacing,

V_f^2-V_i^2 = 2a\Delta X

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Re-arrange for a,

a = -\frac{0.9^2}{2*30*0.3}

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At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

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P = \frac{W}{t} = \frac{4.41}{20}

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goblinko [34]
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e-lub [12.9K]

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Answer:

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