If the resultant force acting on a 2kg object is equal to (3i+4j)N, what is the change in kinetic energy as the object moves from (7i-8j)m to (11i-5j)m?

Question

If the resultant force acting on a 2kg object is equal to (3i+4j)N, what is the change in kinetic energy as the object moves from (7i-8j)m to (11i-5j)m?

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6 answers
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Answer 1

Damon answered
6 years ago

get x motion vector

(11i-5j) - (7i+8j) = 4 i + 3 j

F dot x = (3 i + 4 j) dot ( 4 i + 3 j )
= 12 + 12 = 24 Joules work done
= change in kinetic energy

Answer 2

Fuzail aftab answered
2 years ago

If the resultant on a 2.0 kg object is equal to (3i +4j) N , what is the change in kinetic energy as the object moves from (7i-8j) m

in to (11i-5j)m

Answer 3

Unako answered
1 year ago

|F|=√(3)²+(4)²

=5N

|m|=√(3)²+(4)²
=5m

W=F∆xcos¶
=5(5)cos(0)
=25J

Answer 4

Bot answered
1 year ago

The work done by the resultant force in moving the object from (7i -8j) m to (11i -5j) m is 25 J (Joules).

As per the work-energy principle, the work done by all forces acting on an object is equal to the change in kinetic energy of the object. Therefore, the change in kinetic energy of the 2.0kg object is also 25 J.

Answer 5

Clown Bot answered
1 year ago

Well, let's see here. To calculate the change in kinetic energy, we need to determine the work done on the object. And to calculate the work done, we need to find the dot product of the force and the displacement vectors.

So, first let's calculate the dot product. *Puts on thinking cap* Okay, so the dot product of (3i+4j)N and (11i-5j)m is (3*11) + (4*-5) = 33 - 20 = 13Nm.

Now, since the force applied is perpendicular to the displacement, the work done is zero. And since the work done is zero, the change in kinetic energy is also zero.

Congrats, my friend! The object didn't gain or lose any kinetic energy while traveling from point A to point B. It must have been having a lazy day.

Answer 6

Explain Bot answered
1 year ago

To find the change in kinetic energy, we need to calculate the work done on the object. The work done is defined as the dot product of the force and the displacement vector.

Given:
Mass of the object (m) = 2 kg
Force (F) = 3i + 4j N
Initial position (r1) = 7i - 8j m
Final position (r2) = 11i - 5j m

First, we need to find the displacement vector (Δr) by subtracting the initial position from the final position:
Δr = r2 - r1 = (11i - 5j) - (7i - 8j)
= 11i - 5j - 7i + 8j
= 4i + 3j

Next, we calculate the dot product of the force and displacement vectors:
W = F · Δr = (3i + 4j) · (4i + 3j)
= (3 * 4) + (4 * 3)
= 12 + 12
= 24 N

The work done on the object is 24 N. This work done represents the change in kinetic energy (ΔKE) of the object.

Therefore, the change in kinetic energy as the object moves from (7i - 8j) m to (11i - 5j) m is 24 J (Joules).