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yuradex [85]
3 years ago
7

A 10-kg block is suspended in air by a single string that passes through a pulley and is attached on the other side to a 35-kg b

lock that is on the verge of sliding on the ground. What is the coefficient of static friction between the larger block and the ground? (Note: The magnitude of the vertical force the small block exerts on the large block is 87 N.)
Physics
1 answer:
s2008m [1.1K]3 years ago
4 0

Answer:

0.286

Explanation:

Let g = 10m/s2, and assume the pulley is frictionless, the weight by the 10kg block would exert a force on the 35kg block on the ground

Weight of the 10 kg-block = mg = 10 * 10 = 100 N

This force is balanced by static friction force from the ground due to the normal force of the 35kg block. So the friction force would also be 100N

The weight and normal force of the 35kg block is N = Mg = 35*10 = 350 N

The coefficient of the friction force if friction force divided by normal force:

\mu = \frac{F_f}{N} = \frac{100}{350} = 0.286

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A grocery cart of mass 16 kg is being pushed at a constant speed up a 12-degree ramp by a force of [{MathJax fullWidth='false' f
lawyer [7]

Answer:

If the cart is being pushed at a constant speed, then the acceleration in the direction of motion is zero. Hence, the force in the direction of the motion is zero, according to Newton's Second Law.

\Sigma F_x = ma_x

For simplicity, I will denote the direction along the inclined ramp as x-direction.

In the question the value of the force is not clearly given, so I will denote it as F_P

\Sigma F_{net_x} = ma_x\\\Sigma F = F_{P}\cos(29^\circ) - mg\sin(12^\circ) = ma_x = 0\\F_{P}\cos(29^\circ) = mgsin(12^\circ)\\F_{P}\times 0.8746 = 16\times 9.8\times  0.2079\\F_{P} = 37.2740

Here the angle between the applied force and the x-direction is 12° + 17° = 29°

The x-component of the weight of the cart is equal to sine component of the weight.

Since the cart is rolling on tires the kinetic friction does no work.

Work done by the applied force:

W_{F_P} = F_P_x \cos(29^\circ)\times 7.5 = 244.5 ~J

Work done by the weight of the cart:

W_{mg} = -mg\sin(12^\circ)\times 7.5 = -16\times 9.8 \times 0.2079 \times 7.5 = - 244.5~J

Since the x-component of the weight is in the -x-direction, its work is negative.

Conveniently, the total work done on the particle is zero, since its velocity is constant.

5 0
3 years ago
Force = mass x acceleration
zloy xaker [14]
The answer is (B 60N) hope it helps and have a good day!
4 0
4 years ago
A hill is 132 m long and makes an angle of 12.0 degrees with the horizontal. As a 54 kg jogger runs up the hill, how much work d
TEA [102]

Answer:

14523.55J

Explanation:

The work done by the jogger against gravity is given by the following equation;

W=mgh.................(1)

where m is the mass, g is acceleration due to gravity taken as 9.8m/s^2 and h is the height of the hill.

Since the length of the hill is 132m and it is inclined at 12 degrees to the horizontal, the height is thus given as follows;

h=132sin12^o\\h=27.44m

Substituting this into equation (1) with all other necessary parameters, we obtain the following;

W=54*9.8*27.44\\W=14523.55J

4 0
3 years ago
Different types of heat transfer: What is radiation?
denpristay [2]
The transfer of internal energy in the form of electromagnetic waves.
3 0
4 years ago
Two blocks, stacked one on top of the other, slide on a frictionless horizontal surface. The surface between the two blocks is r
devlian [24]

Answer:

F_{net} = 31.88 N

Explanation:

When top block is just or about to slide on the lower block then we can say that the frictional force on it will be maximum static friction

So we will have

F_{net} = ma

F_s = ma

\mu_s mg = ma

a = \mu_s g

a = (0.50)(9.81)

a = 4.905 m/s^2

now for the Net force on two blocks to move together

F_{net} = (m_1 + m_2) a

F_{net} = (2.3 + 4.2)(4.905)

F_{net} = 31.88 N

4 0
3 years ago
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