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ELEN [110]
3 years ago
14

Suppose you have a 136-kg wooden crate resting on a wood floor. The coefficient of static friction is 0.4 and coefficient of kin

etic friction is 0.2 What maximum force (in N) can you exert horizontally on the crate without moving it? (Hint: static friction)
Physics
1 answer:
stealth61 [152]3 years ago
4 0

Answer:

The the maximum force acting on the crate is 533.12 newtons.

Explanation:

It is given that,

Mass of the wooden crate, m = 136 kg

The coefficient of static friction, \mu_s=0.4

The coefficient of kinetic friction, \mu_k=0.2

We need to find the maximum force exerted horizontally on the crate without moving it. As the crate is not moving than the coefficient of static friction will act and the force is given by :

F=\mu_s mg

F=0.4\times 136\ kg\times 9.8\ m/s^2

F = 533.12 N

So, the maximum force acting on the crate is 533.12 newtons. Hence, this is the required solution.

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As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

9.8\times1.94=\dfrac{1}{2}\times v^2

v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

6 0
3 years ago
Philosophy: The Big Picture Unit 8
SVEN [57.7K]

D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.
5 0
3 years ago
The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.
AnnyKZ [126]

Answer:

F = 2 * 30 / 5 = 12 N to stop forward motion

F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees

(12^2 + 16^2)^1.2 = 20 N   average force applied

5 0
3 years ago
The kinetic energy K of a moving object varies jointly with its mass m and the square of its velocity v. If an object weighing 4
nata0808 [166]

Answer:

K' = 1777.777  J

Explanation:

Given that

m = 40 kg

v= 15 m/s

K=1000

Given that kinetic energy(K) varies with mass(m) and velocity(v)

 K= C(mv²)

Where

C= Constant

m=mass

v=velocity

When

m = 40 kg ,v= 15 m/s ,K=1000

 K= C(mv²)

1000 = C( 40 x 15²)

C=0.111111

When m = 40 kg and v= 20 m/s

K' = C(mv²)

K= 0.1111 x (40 x 20²)

K' = 1777.777  J

5 0
3 years ago
Read 2 more answers
A ball is thrown horizontally from the top of a building 21.8 m high. The ball strikes the ground at a point 101 m from the base
riadik2000 [5.3K]

Answer:

t=2.10 s

u= 47.40 m/s

Explanation:

given that

h= 21.8 m

x= 101 m

g=9.8 m/s²

Lets take horizontal speed of  ball = u m/s

The vertical speed of the car at initial condition is zero ( v= 0).

We know that

h=vt+\dfrac{1}{2}gt^2

v= 0 m/s

h=\dfrac{1}{2}gt^2

now by putting the values

21.8 = 1/2 x 9.8 x t²

t=2.10 s

This is time when ball was in motion.

Now in horizontal direction

x = u .t

101 = u x 2.1

u= 47.40 m/s

6 0
3 years ago
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