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Dmitry_Shevchenko [17]

3 years ago

6

A 35 kg boy is riding a 65 kg go-cart. He pushes on the gas pedal, causing the cart to accelerate at 5 m/s2. Use the equation F

= m × a to find the force exerted by the go-cart’s engine.

2 answers:

trasher [3.6K]3 years ago

4 0

Answer:

$\boxed {\boxed {\sf 500 \ Newtons }}$

Explanation:

The equation for force is given:

$F=m*a$

First, we must find the total mass, which is the sum of the boy's mass and the go-cart's mass.

total mass= boy's mass + go cart's mass

The boy's mass is 35 kilograms and the go cart's is 65 kilograms.

total mass= 35 kg+ 65 kg=100 kg

Now we know the total mass and the acceleration.

$m= 100 \ kg \\a= 5 \ m/s^2$

Substitute the values into the formula.

$F=100 \ kg * 5 \ m/s^2$

Multiply.

$F= 500 \ kg*m/s^2$

1 kilograms meter per square second is equal to 1 Newton.
Our answer of 500 kg*m/s² is equal to 500 Newtons.

$F= 500 \ N$

The force exerted by the go cart engine is <u>500 Newtons.</u>

amid [387]3 years ago

4 0

Answer:

500 N

Explanation:

Your Welcome. (:

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A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

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Explanation:

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