PLEASE SHOW FULL STEPS Problem 1 : consider a block of mass 5 kg on a horiztonal

Question

PLEASE SHOW FULL STEPS

Problem 1 :

consider a block of mass 5 kg on a horiztonal surface. A pulling force is directed at an angle of 30 degree above the horitonal as shown in the figure below. the coefficient of static friction between the surface and the blcok is 0.4 and the coefficient of kinetic fraction between the surface and the block is 0.25

a) if the pulling force has a size of 20 N and the surface is frictionless, find the acceleration of the block

b) if the pulling force has a size of 50 N and the surface is frictionless, find the acceleration of the block

problem 2 : consider the ste up shown below. Two masses m1= 4 kg and m2= 2kg are connected by a massless rope and a frictionless pulley as shown. Find the tension in the rope and the acceleration of the blocks if:

a) the table surface is frictionaless

b) the coefficients of friction betwwen the table and m1 are

consider a block of mass 5 kg on a horizontal surface. A pulling force is directed at an angle of 30 degree above the horizontal as shown in the figure below. the coefficient of static friction between the surface and the blcok is 0.4 and the coefficient of kinetic fraction between the surface and the block is 0.25 if the pulling force has a size of 20 N and the surface is frictionless, find the acceleration of the block if the pulling force has a size of 50 N and the surface is frictionless, find the acceleration of the block Consider the ste up shown below. Two masses m1= 4 kg and m2= 2kg are connected by a massless rope and a frictionless pulley as shown. Find the tension in the rope and the acceleration of the blocks if: the table surface is frictionaless the coefficients of friction between the table and m1 are mu S = 0.2, mu K = 0.15. the coefficients of friction between the table and m1 are mu S = 0.8, mu K = 0.6 Consider the two blocks connected by a massless rope shown below. A second massless rope is attached to the first block. A force Fp is applied to another rope in order to accelerate the two block system with an acceleration of 5 m/s^2 upward. Take m2= 4kg and m1= 2kg. Do the following parts in whatever order you find convenient. Find Fp. Find the force that rope 2 exerts on m1. Find the force that rope 2 exerts on m2. Find the force that rope 1 exerts on m1. repeat the problem if each rope has a mass of 0.25 kg

Explanation / Answer

1a)

Weight = mg = 5*9.8 = 49

Normal force = N = mg - Fsin30 = 49 - 20*0.5 = 39

horizontal force = Fcos30 = 17.32

friction force static = uN = 0.4*39 = 15.6 this is less than horizontal force

hence block will move

friction force kinetic = uN = 0.25*39 = 9.75

acceleration = force/mass = 17.32-9.75 / 5 = 1.514

1b)

Weight = mg = 5*9.8 = 49

Normal force = N = mg - Fsin30 = 49 - 50*0.5 = 24

horizontal force = Fcos30 = 43.3

friction force static = uN = 0.4*24 = 9.6 this is less than horizontal force

hence block will move

friction force kinetic = uN = 0.25*24 = 6

acceleration = force/mass = 43.3-6 / 5 = 7.46

2a)

let acceleration of blocks be a

m2g-T=m2a

T=m1a

hence

m2g - m1a = m2a

hence

a = 3.27

T = 13.1

2b)

Normal force on block 1 = m1g = 39.2

static friction force = uN = 0.2*39.2 = 7.84

kinetic friction force = uN = 0.15*39.2 = 5.88

now

m2g-T=m2a

T- 5.88 = m1a

hence

T = 15.04

a = 2.29

2c)

Normal force on block 1 = m1g = 39.2

static friction force = uN = 0.8*39.2 = 31.36 is less than tension

kinetic friction force = uN = 0.6*39.2 = 23.52

now

m2g-T=m2a

T- 23.52 = m1a

hence

T = 33.97

a = 2.61

3a)

Fp - (m1g+m2g)= (m1+m2)a

Fp = 6g + 6a = 6*9.8 + 6*5 = 88.8

3b)

Fp - T = m1a

88.8 - T = 2*5

T = 78.8 downward

3c)

T = 78.8 upward

3d)

Fp = 88.8 upward

3e)

Fp - (m1+m2 + 0.25 + 0.25)g = (m1+m2+ 0.25+0.25)a

Fp = 96.2

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