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belka [17]
3 years ago
6

Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the

direction of travel. Explain what your answer means in terms of the object’s energy.
Physics
1 answer:
aivan3 [116]3 years ago
7 0

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, \theta=135^{\circ}

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

W=Fd\ cos\theta

W=50\times 9\times \ cos(135)

W = -318.19 Joules

So, the work done by the force is 318.19 Joules. The work is done in opposite to the direction of motion. Hence, this is the required solution.

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Is Mia right and did you show a lot of work?​
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square root A 1400 kg car is coasting on a horizontal road with a speed of 18 m/s . After passing over an unpaved, sandy stretch
Lina20 [59]

Answer:

The net force on the car is 2560 N.

Explanation:

According to work energy theorem, the amount of work done is equal to the change of kinetic energy by an object. If 'W' be the work done on an object to change its kinetic energy from an initial value 'K_{i}' to the final value 'K_{f}', then mathematically,

W = K_{f} - K_{i} = \dfrac{1}{2}~m~(v_{f}^{2} - v_{i}^{2})........................................(I)

where 'm' is the mass of the object and 'v_{i}' and 'v_{f}' be the initial and final velocity of the object respectively. If 'F_{net}' be the net force applied on the car, as per given problem, and 's' is the displacement occurs then we can write,

W = F_{net}~.~s.......................................................(II)

Given, m = 1400~Kg,~v_{i} = 18~m~s^{-1}~v_{f} = 14~m~s^{-1}~and~s = 35~m.

Equating equations (I) and (II),

&& - F_{net} \times 35~m = \dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})~m^{2}~s^{-2}\\&or,& F_{net} = \dfrac{\dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})}{35}~N\\&or,& F_{net} = 2560~N

6 0
3 years ago
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