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AleksAgata [21]

3 years ago

7

In order to increase the amount of work completed, it is necessary to decrease the force applied to an object. decrease the time

to do the work. increase the time to do the work. increase the force applied to the object.

2 answers:

Misha Larkins [42]3 years ago

8 0

Increase force applied to object

wlad13 [49]3 years ago

5 0

<span> <span> The answer to your question is: increase the force applied to the object.

Two items are provided as a basis for that conclusion:
1. According to Newton's Second Law of Motion, the formula for finding force is: F = ma
where F is the force,
m is the mass of an object,
and a is the acceleration of the object.

And 2: work = force x distance or W = F x d.</span></span>

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Two constant forces act on an object of mass m = 5.60 kg object moving in the xy plane as shown in the figure below. Force F1 is

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Answer:

Part a)

$F_1 = 22.5\hat i + 15.7 \hat j$

$F_2 = -35.5 \hat i + 20.5 \hat j$

Part b)

$F = -10 \hat i + 36.2 \hat j$

Part c)

$a = -1.78 \hat i + 6.46 \hat j$

Part d)

$v_f = -0.64 \hat i + 21.93\hat j$

Part e)

$r_f = 6.09\hat i + 36.72 \hat j$

Part f)

$KE = 1347.7 J$

Part g)

$KE = 1348 J$

Explanation:

Part a)

first force is given as

$F_1 = 27.5 N$ at 35 degree

$F_1 = 27.5 cos35 \hat i + 27.5 sin35 \hat j$

$F_1 = 22.5\hat i + 15.7 \hat j$

Second force is given as

$F_2 = 41 N$ at 150 degree

$F_2 = 41 cos150 \hat i + 41 sin150\hat j$

$F_2 = -35.5 \hat i + 20.5 \hat j$

Part b)

Total force is given as

$F = F_1 + F_2$

$F = (22.5 - 35.5)\hat i + (15.7 + 20.5)\hat j$

$F = -10 \hat i + 36.2 \hat j$

Part c)

as we know

F = ma so object acceleration is given as

$a = \frac{F}{m}$

$a = \frac{-10}{5.60} \hat i + \frac{36.2}{5.60} \hat j$

$a = -1.78 \hat i + 6.46 \hat j$

Part d)

By kinematics we know that

$v_f = v_i + at$

$v_f = (4.70 \hat i + 2.55 \hat j) + (-1.78\hat i + 6.46\hat j)(3)$

$v_f = -0.64 \hat i + 21.93\hat j$

Part e)

As we know that final position is given as

$r_f = v_i t + \frac{1}{2}at^2$

$r_f = (4.70 \hat i + 2.55 \hat j)(3) + \frac{1}{2}(-1.78 \hat i + 6.46 \hat j)(3^2)$

$r_f = 6.09\hat i + 36.72 \hat j$

Part f)

final kinetic energy is given as

$KE = \frac{1]{2}mv_f^2$

$KE = \frac{1}{2}(5.60)(0.64^2 + 21.93^2)$

$KE = 1347.7 J$

Part g)

final kinetic energy is given as

$KE = KE_i + F.r$

$KE = \frac{1}{2}(5.60)(4.70^2 + 2.55^2) + (-10 \hat i + 36.2 \hat j).(6.09\hat i + 36.72\hat j)$

$KE =80 + 1268.4$

$KE = 1348 J$

4 0

3 years ago