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1 year ago

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A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction actin

g on the box is a constant 20 N. How much work is done by the 50 N force?

1 answer:

Stels [109]1 year ago

8 0

The work is done by the 50 N force is 200 J.

<h3>Work done by the 50 N force</h3>

Work done = Fd

where;

F is Force applied
d is the displacement of the object

Work done = 50 N x 4 m = 200 J

Thus, the work is done by the 50 N force is 200 J.

Learn more about work done here: brainly.com/question/8119756

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Calcium-51 has a half-life of 4.5 days. Only 0.75 gram remains of a sample that was initially 12 grams. How old is the sample of

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How many times did the original sample lose 50% of its radioactivity ?

-- Start with. . . . . . . . . . . . 12 grams.
-- Lose half of it once. . . . . . 6 grams left.
-- Lose half of it again . . . . . 3 grams left.
-- Lose half of it again . . . . . 1.5 grams left.
-- Lose half of it again . . . . . 0.75 gram left.

-- How many times did it lose half ? 4 times.

-- How long does it take to lose half ? 4.5 days.
(That's why it's called the 'half-life'.)

-- How long did it take to lose half, 4 times ?

(4 x 4.5 days) = 18 days .

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2 years ago

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Answer:

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Explanation:

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2 years ago

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

6 0

3 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago