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A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance

Talja [164]

Answer:

21 m

Explanation:

Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.

Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N

The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N

The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N

If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.

Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.

For the block to be in contact with the surface, the vertical forces on the block must balance.

Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,

N = mg + F₃" (since both the weight and the resultant vertical force act downwards)

N = mg + F₃"

Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N

So,

N = mg + F₃"

N = 35 kg × 9.8 m/s² + 8.82 N

N = 343 N + 8.82 N

N = 351.82 N

So, the net horizontal force F = F₃' - f.

F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

4 0

3 years ago

In a medical lab, Sandrine is working to isolate one element from a sample of liquid material. She uses a centrifuge, a machine

Ludmilka [50]

Answer:

1

Explanation:

4 0

2 years ago

If you are being tailgated and need to make a stop, you should

alexandr402 [8]

If you are being tailgated and need to make a stop, you should:

B. flash your brake lights ahead of time.

C. slow sooner to make a gradual stop.

<h3>What is tailgating?</h3>

Tailgating can be defined as an act which typically involves a tailgater driving dangerously close behind another vehicle that is in front.

This ultimately implies that, a tailgating vehicle is a hazard to any driver because when he or she brakes suddenly, it may result in the tailgater hitting him or her from the rear.

In this context, we can infer and logically deduce that if you are being tailgated and need to make a stop, you should flash your brake lights ahead of time and slow sooner to make a gradual stop.

Read more on tailgating vehicle here: brainly.com/question/14287208

#SPJ1

8 0

2 years ago

How can the use of local materials improve the standard of living of Filipinos?

Rudiy27

Choose a quality one, and don't use it as necessary

7 0

2 years ago

Refrigerant 134a at p1 = 30 lbf/in2, T1 = 40oF enters a compressor operating at steady state with a mass flow rate of 200 lb/h a

Mazyrski [523]

Answer:

a) $\mathbf{Q_c = -3730.8684 \ Btu/hr}$

b) $\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}$

Explanation:

From the properties of Super-heated Refrigerant 134a Vapor at $T_1 = 40^0 F$ , $P_1 = 30 \ lbf/in^2$ ; we obtain the following properties for specific enthalpy and specific entropy.

So; specific enthalpy $h_1 = 109.12 \ Btu/lb$

specific entropy $s_1 = 0.2315 \ Btu/lb.^0R$

Also; from the properties of saturated Refrigerant 134 a vapor (liquid - vapor). pressure table at $P_2 = 160 \ lbf/in^2$ ; we obtain the following properties: