
(a)
The work done by the force from
(a)

Explanation of Solution
Given:
Mass of the object is
Velocity of the object is
Formula used:
Write expression for work done by the force from
Here,
Write expression for area of square.
Here,
Calculation:
Substitute
Substitute
Conclusion:
Thus, work done by the force is
(b)
The kinetic energy of object at
(b)

Explanation of Solution
Given:
Mass of the object is
Velocity of the object is
Formula used:
Write expression for work done by the force from
Here,
Write expression for area of square.
Here,
Write expression for kinetic energy at
Substitute
Calculation:
Substitute
Substitute
Substitute
Conclusion:
Thus, kinetic energy at
(c)
The speed of object at
(c)

Explanation of Solution
Given:
Mass of the object is
Velocity of the object is
Formula used:
Write expression for velocity at
Calculation:
Substitute
Conclusion:
Thus, the velocity of object is
(d)
The work done on the object from
(d)

Explanation of Solution
Given:
Mass of the object is
Velocity of the object is
Formula used:
Write expression for work done on the object from
Here,
Write expression for area of square.
Here,
Calculation:
Substitute
Substitute
Conclusion:
Thus, work done is
(e)
The speed of object at
(e)

Explanation of Solution
Given:
Mass of the object is
Velocity of the object is
Formula used:
Write expression for kinetic energy at
Substitute
Write expression for velocity at
Calculation:
Substitute
Substitute
Conclusion:
Thus, velocity of the object is
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Chapter 6 Solutions
Physics for Scientists and Engineers
- An object is placed 24.1 cm to the left of a diverging lens (f = -6.51 cm). A concave mirror (f= 14.8 cm) is placed 30.2 cm to the right of the lens to form an image of the first image formed by the lens. Find the final image distance, measured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?arrow_forwardConcept Simulation 26.4 provides the option of exploring the ray diagram that applies to this problem. The distance between an object and its image formed by a diverging lens is 5.90 cm. The focal length of the lens is -2.60 cm. Find (a) the image distance and (b) the object distance.arrow_forwardPls help ASAParrow_forward
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