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3 years ago

"A window washer pulls herself upward using the bucket-pulley apparatus. The mass of the person plus the bucket is 65 kg.

Physics

2 answers:

FromTheMoon [43]3 years ago

8 0

Part A.

If there is a constant speed, therefore the net force is zero.

Forces acting are the:

1. Weight

2. Upward force exerted by the rope which is opposite in direction but the magnitude is to the force which is acting downwards exerted on the rope.

∑Force = pulling force - weight = 0

Weight = mg

Hence,

F (pulling force) = mg

Substituting the values,

F (pulling force) = 65 × 9.81

F (pulling force) = 637.65 Newton

____________________________________________________________

Part B.

F = ma

10% increase in the force = 1.10 ₓ 637.65

10% increase in the force = 701.415

Therefore,

701.415 = 65 ₓ a

a = 701.45/65

a = 10.8 m/s²

Aleksandr [31]3 years ago

7 0

Consider the free body diagrams as shown in the file attached.

Let the acc of system is a. 
For the washer T+N-m1g=m1a ............................(1)
For the bucket T-N-m2g=m2a ............................(2)

(A)
a=0 Eqns (1) and (2) become T+N-m1g=0 T-N-m2g=0 Adding and rearranging, T=(m1+m2)/2g T=325N 
(B)
 We are specified the value of T here as 325+32.5=357.5N eqns (1)and (2) become 357.5+N-m1g=m1a 357.5-N-m2g=m2a Adding, 715-(m1+m2)g=(m1+m2)a

Solve for 'a'

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A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the

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Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

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$s_1$ is the displacement of the car in the 1st part

Solving for $s_1$ ,

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We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$

where:

$v_1=32.4 m/s$ is the initial velocity at the beginning of the 2nd phase

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$t_2=5 s$ is the time elapsed in the 2nd phase

Substituting,

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So, the total displacement of the car is

$s=s_1+s_2=324+81=405 m$

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