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kobusy [5.1K]
3 years ago
12

A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.

0 N is applied to the block in the horizontal direction (not parallel to the ramp).a) If this force is just large enough to start the block moving down the ramp, what is the coefficient of static friction μsbetween the block and the ramp?b) If the coefficient of kinetic friction between the block and the ramp has a value of 0.40, what is the magnitude of the acceleration of the block along the ramp as this force is applied to the block?c) What is the work done by this applied force if the block travels a distance of 2.0 m down (i.e., parallel to) the ramp?
Physics
1 answer:
Marina86 [1]3 years ago
8 0

Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

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A ball thrown with 50N of force accelerates at 25 m/s2.  What is the mass of the ball?
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F=ma
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7 0
4 years ago
A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12.0 m/s. What minimum coefficient of friction must
Sergio039 [100]

Explanation:

The centripetal force F_c on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.

y:\:\:\:\:N - mg = 0

x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N

or

m \dfrac{v^2}{r} = \mu mg

Solving for \mu,

\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28

3 0
3 years ago
A charge Q is distributed uniformly around the perimeter of a ring of radius R. Determine the electric potential difference betw
il63 [147K]

Answer:

the electric potential difference between the point at the center of the ring and a point on its axis ΔV is ( 0.8356 )\frac{kQ}{R}

Explanation:

Given the data in the question;

electric potential at the center of the ring V₀ = kQ / R

electric potential on the axis point Vr = kQ / √( R² + x² )

at a distance 6R from the center,

point at x = 6R

so distance circumference r = √( R² + (6R)² )

so

electric potential on the axis point Vr = kQ / √( R² + (6R)² )

Vr = kQ / R√37

Now

ΔV = V₀ - Vr

we substitute

ΔV = ( kQ / R) - ( kQ / R√37 )

ΔV =  kQ/R( 1 - 1/√37 )

ΔV =  kQ/R( 1 - 0.164398987 )

ΔV =  kQ/R( 0.8356 )

ΔV = ( 0.8356 )\frac{kQ}{R}       { where k = \frac{1}{4\pi e_0} }

Therefore, the electric potential difference between the point at the center of the ring and a point on its axis ΔV is ( 0.8356 )\frac{kQ}{R}

5 0
3 years ago
1. A person of mass 55 kg is in a bumper car that has a mass of 75 kg. They are together at a velocity of 8 m/s.
ExtremeBDS [4]

The momentum of the bumper car after the collision is 1,040 kgm/s.

<h3>Momentum of bumper</h3>

The change in momentum of the bumper is calculated as follows;

P = v(m1 + m2)

P = 8(55 + 75)

P = 1,040 kgm/s

The momentum of the bumper car after the collision is 1,040 kgm/s.

The direction is still the same.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

7 0
2 years ago
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