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tekilochka [14]

2 years ago

13

A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car

moves 30.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car.
(a) the speed v

______m/s

(b) the horizontal force exerted on the car (Enter the magnitude.)
_______N

1 answer:

Illusion [34]2 years ago

5 0

The horizontal force applied is 160 N while the velocity is 2.03 m/s.

<h3>What is the speed of the car?</h3>

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

Learn more about force and work:brainly.com/question/758238

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A man pushed a cabinet with a force of 200N. What is the mass of the cabinet that accelerates 4 m/s/s?

ELEN [110]

Answer:

$\boxed {\boxed {\sf 50 \ kg}}$

Explanation:

We are asked to find the mass of a cabinet, given the force and acceleration. According to Newton's Second Law of Motion, force is the product of mass and acceleration. The formula for this is:

$F= m \times a$

The force is 200 Newtons, but we should convert the units to make unit cancellation easier. 1 Newton is equal to 1 kilogram meter per second squared, so the force of 200 Newtons is 200 kilogram meters per second squared.

The mass is unknown and the acceleration is 4 meters per second per second or 4 meters per second squared.

F= 200 kg*m/s²
a= 4 m/s²

Substitute the values into the formula.

$200 \ kg *m/s^2 = m \times 4 \ m/s^2$

We are solving for the mass, m, so we must isolate the variable. It is being multiplied by 4 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 4 m/s²

$\frac {200 \ kg *m/s^2}{4 \ m/s^2}= \frac{m \times 4\ m/s^2}{4 \ m/s^2}$

$\frac {200 \ kg *m/s^2}{4 \ m/s^2} =m$

The units of meters per second squared cancel.

$\frac {200 \ kg }{4 }=m$

$50 \ kg =m$

The mass of the cabinet is <u>50 kilograms.</u>

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3 years ago

A box sits on the back of a flatbed truck. If the coefficient of static friction between the box and the truck bed is μs = 0.400

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Answer:

28,699m

Explanation:

The force to make the box move should be <u>μs.N=μs.m.g=m.</u><u>|</u><u>a</u><u>|</u>

then,

|a|=μs.g

Being

μs coefficient of static friction,

N the force made by the truck on the box caused by the gravity force,

m the mass,

g the acceleration of gravity

and a the acceleration of the truck.

$x = v \times t + \frac{1}{2} \times a \times {t}^{2}$

as the truck is stopping, the acceleration is negative. then,

$x = v \times t - \frac{1}{2} \times |a| \times {t}^{2}$

$|a| = v \div t \\ t = v \div |a|$

$x = v \times (v \div |a| ) - \frac{1}{2} \times |a| \times {(v \div |a|)}^{2}$

$x = v \times (v \div μs.g ) - \frac{1}{2} \times |a| \times {(v \div μs.g)}^{2}$

$x = \frac{{v}^{2}}{μs \times g} - \frac{1}{2} \times \frac{{v}^{2}}{μs \times g} \\ x = \frac{1}{2} \times \frac{{v}^{2}}{μs \times g} \\ x = 0.5 \times \frac{{(15m/s) }^{2} }{0.4 \times 9.8m/ {s}^{2} } = 28.699m$

28,699m

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A 0.0382-kg bullet is fired horizontally into a 3.78-kg wooden block attached to one end of a massless, horizontal spring (k = 8

wel

Answer:

$280.87$ ms⁻¹

Explanation:

Consider the motion of the bullet-block combination after collision

$m$ = mass of the bullet = 0.0382 kg

$M$ = mass of wooden block = 3.78 kg

$V$ = velocity of the bullet-block combination after collision

$k$ = spring constant of the spring = 833 N m⁻¹

$A$ = Amplitude of oscillation = 0.190 m

Using conservation of energy

Kinetic energy of bullet-block combination after collision = Spring potential energy gained due to compression of spring

$(0.5)(m + M)V^{2} = (0.5)kA^{2}$

$(0.0382 + 3.78)V^{2} = (833)(0.190)^{2}$

$V = 2.81$ ms⁻¹

$v_{o}$ = initial velocity of the bullet before striking the block

Using conservation of momentum for the collision between bullet and block

$m v_{o} = (m + M) V$

$(0.0382) v_{o} = (0.0382 + 3.78) (2.81)$

$v_{o} = 280.87$ ms⁻¹

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