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Bogdan [553]
3 years ago
7

An initially stationary 4.3 kg object accelerates horizontally and uniformly to a speed of 11 m/s in 3.4 s. (a) In that 3.4 s in

terval, how much work is done on the object by the force accelerating it
Physics
2 answers:
Nataliya [291]3 years ago
8 0

Answer:

The work done on the object by the force accelerating it is 520.31 J

Explanation:

Given;

mass of an object = 4.3 kg

horizontal velocity of the object, v = 11 m/s

time of acceleration, t = 3.4 s

work done is given as the product of force and distance

Work done = Fd

horizontal distance traveled by the object within 3.4 s, is calculated as follows;

X = Vt + ¹/₂gt², gravity has little or no influence on horizontal displacement, thus g = 0

X = Vt

X = 11*3.4 = 37.4 m

Force on the object, F = ma = m(v/t) = 4.3(11/3.4) = 13.912 N

work done = Fd = 13.912 x 37.4 = 520.31 J

Therefore, the work done on the object by the force accelerating it is 520.31 J

kompoz [17]3 years ago
3 0

Answer: W=260.174J

Explanation: since the object is stationary, it initial velocity U = 0

final velocity V = 11 m/s, time t = 3.4s, distance S = ?, acceleration a = ? work done W = ? force F = ?

W = FS

a = v-u/t = 11-0/3.4 = 3.2353m/s^2

to calculate the distance, let look at one of the equations of motion

V^2=U^2+2as hence s = V^2-U^2/2a = 11*11/2*3.235 = 18.7017m

But force F = MA (mass*acceleration)

F= 4.3*3.2353 = 13.91179N

therefore work done W = 13.91179*18.7017 = 260.174J

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Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
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Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

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     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

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    T = m_A a

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We solve this system of equations

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    a = m_B / (m_A + m_B) g

In this initial case

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     m_B = M

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Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

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We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

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