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tekilochka [14]
3 years ago
13

Using a horizontal force of 60 N, a wagon is pushed horizontally across the floor a distance of 12 meters at a constant speed of

2.3 m/s
a) What is the work done by the force?
b) What is the power supplied by the force?
Physics
1 answer:
RideAnS [48]3 years ago
3 0

Answer:

a) The work done by the force is 720 joules, b) The power supplied by the force is 138 watts.

Explanation:

a) Since force is uniform and parallel to the direction of motion, the work (W), in joules, done by the force (F), in newtons, is defined by this formula:

W = F\cdot s (1)

Where s is the travelled distance, in meters.

If we know that F = 60\,N and s = 12\,m, then the work done by the force is:

W = F\cdot s

W = (60\,N)\cdot (12\,m)

W = 720\,J

The work done by the force is 720 joules.

b) And an expression for the power supplied by the force (\dot W), in watts, is concieved by differentiating (1) in time:

\dot W = F\cdot \dot s

Where \dot s is the speed of the wagon, in meters per second.

If we know that F = 60\,N and \dot s = 2.3\,\frac{m}{s}, then the power supplied by the force is:

\dot W = F\cdot \dot s

\dot W = (60\,N)\cdot \left(2.3\,\frac{m}{s} \right)

\dot W = 138\,W

The power supplied by the force is 138 watts.

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this is a 3 part questionOn vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85
mamaluj [8]

ANSWER:

(a) 1036 N

(b) -1036 N

(c) 2590 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}

(b)

\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}

(c)

\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}

3 0
1 year ago
A car starts from rest and accelerates uniformly for a five seconds along a straight road. If speed obtained by the car is 72 km
Step2247 [10]

Answer:

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Explanation:

Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.

v_f=v_o+at

Since the car starts at rest, you can write the following equation:

20=0+a(5) \\\\a=20\div 5=4 m/s^2

Now that you have the acceleration, you can do this:

d=v_o+\dfrac{1}{2}at^2

Once again, there is no initial velocity:

d=\dfrac{1}{2}(4)(5)^2=2 \cdot 25=50m

Hope this helps!

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