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dalvyx [7]
3 years ago
7

Mr. Covault gave his students an assignment to design and conduct experiments that would allow them to find the relationship bet

ween force, mass, and acceleration. Some students decided to use a setup like the one below:
The applied force in this setup is equal to the weight of the item attached to the end of the pulley. One student tested the acceleration due to this force on a wooden block (as shown above); another student used a wheeled cart instead of a wooden block.

Mr. Covault's other students decided to simply drop objects. The force in that case is the weight of the object.

All the students measured the time it took each object, starting from rest, to move a certain distance. They used the time and distance to calculate the average acceleration.

Four students' results are shown below. A successful experiment should eliminate all forces acting on the object except the force being investigated. The experiment should confirm Newton's second law:



One newton (N) is 1 kg·m/s2.

Which of the following students had a well designed and conducted experiment?

Student Mass
(kg) Force
(N) Measured
Acceleration
(m/s2)
Kira 0.21 0.098 0.0047
Sophie 0.11 0.098 0.88
Jacques 0.050 0.49 4.9
Chase 0.50 4.9 8.9
Physics
1 answer:
zhenek [66]3 years ago
7 0

Answer:

Sophie

Explanation:

Only Sophie's results supported Newton's second law. The other students' measured accelerations were significantly lower than expected. This could indicate that their experimental designs had not sufficiently eliminated drag forces.

F=ma         Rearranging when solving for acceleration gives:  a=F/m

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Answer:

(a)

x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km

(b) r=41.32km

\alpha =24.23^o

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies 45^o south of east. The y-component will be negative and the x-component will be positive.

x_1=25cos45^o=17.68km

y_1=-25sin45^o=-17.68km

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

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y_2=40sin60^o=34.64km

Therefore, total displacements:

x=x_1+x_2=(17.68+20)km=37.68km

y=y_1+y_2=(-17.68+34.64)km=16.96km

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a. 25 joules of work is done by the object.

Explanation:

According to the work-energy theorem, the work done ON an object is equal to the change in kinetic energy of the object, therefore:

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K_f = \frac{1}{2}mv^2 is the final kinetic energy of the car, with

m = 2 kg being the mass of the car

v = 0 is the final speed of the car (brought to rest)

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the car, with

u = 5 m/s being the initial speed of the car

Substituting numbers, we find:

W=\frac{1}{2}(2)(0)^2 - \frac{1}{2}(2)(5)^2=-25 J

The negative sign means that the work done ON the car is negative: this means therefore that the work has been actually done BY the car. In fact, the car has lost its kinetic energy: this means that it has done work on the surrounding, converting its kinetic energy into other forms of energy.

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