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3 years ago

If you want to lift a 5-kg box to a height of 2 meters, how much work must be

Physics

2 answers:

kramer3 years ago

7 0

Answer:100joules

Explanation:

Work= force×distance

Work=mass×acceleration×distance

Work=5×10×2

Work=100joules

grandymaker [24]3 years ago

6 0

Answer:

98J is the correct answer

Explanation:

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A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo

Taya2010 [7]

Answer:

Net force on the block is 32 N.

Acceleration of the object is 6.4 m/s².

Explanation:

Let the acceleration of the object be $a$ m/s².

Given:

Mass of the block is, $m=5\ kg$

Force of pull is, $F=40\ N$

Frictional force on the block is, $f=8\ N$

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

$F_{net}=F-f=40-8=32\ N$

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

$F_{net}=ma\\32=5a\\a=\frac{32}{5}=6.4\ m/s^2$

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².

3 0

3 years ago

A 7 kg ball is moving at a constant speed of 5 m/s. A force of 300 N is applied to the ball for 4 s. The new speed of the ball i

olasank [31]

Answer:

The new speed of the ball is 176.43 m/s

Explanation:

Given;

mass of the ball, m = 7 kg

initial speed of the ball, u = 5 m/s

applied force, F = 300 N

time of force action on the ball, t = 4 s

Apply Newton's second law of motion;

$F = ma = \frac{m(v-u)}{t}\\\\m(v-u) = Ft\\\\v-u = \frac{Ft}{m}\\\\v = \frac{Ft}{m} + u$

where;

v is new speed of the ball

$v = \frac{Ft}{m} + u\\\\v =\frac{300*4}{7} + 5\\\\v = 176.43 \ m/s$

Therefore, the new speed of the ball is 176.43 m/s

8 0

3 years ago

An 3.7 lb hammer head, traveling at 5.8 ft/s strikes a nail and is brought to a stop in 0.00068 s. The acceleration of gravity i

CaHeK987 [17]

Answer:

31677.2 lb

Explanation:

mass of hammer (m) = 3.7 lb

initial velocity (u) = 5.8 ft/s

final velocity (v) = 0

time (t) = 0.00068 s

acceleration due to gravity (g) 32 ft/s^{2}

force = m x ( a + g )

where

m is the mass = 3.7 lb
g is the acceleration due to gravity = 32 ft/s^{2}
a is the acceleration of the hammer

from v = u + at

a = (v-u)/ t

a = (0-5.8)/0.00068 = -8529.4 ( the negative sign showa the its decelerating)

we can substitute all required values into force= m x (a+g)

force = 3.7 x (8529.4 + 32) = 31677.2 lb

4 0

2 years ago

If you push on a wall with a force of 200 N, with what force does the wall push back?

PSYCHO15rus [73]

200 N, that is if the force is balanced and the wall doesn't move

8 0

2 years ago

How far does a car travel in 90 seconds if it’s traveling 55 m/s? Show equation

podryga [215]

You just said the car is traveling at the speed of 55 m/s. If I understand this correctly, that means the car will cover:

55 meters in the first second,

55 meters in the 2nd second,

55 meters in the 3rd second,

55 meters in the 4th second,

55 meters in the 5th second,

55 meters in the 87th second,

55 meters in the 88th second,

55 meters in the 89th second, and

55 meters in the 90th second.

That's 55 meters 90 times. If you just move these words around a little bit, it says "90 times 55 meters" . . . a pretty simple arithmetic problem.

The equation is . . . <em>Distance = (55 m/s) times (time, in seconds)</em> .

I get <em>4,953 meters</em>. You should check me on this.

8 0

3 years ago