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
(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
in to (11i-5j)m
=5N
|m|=√(3)²+(4)²
=5m
W=F∆xcos¶
=5(5)cos(0)
=25J
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.
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.
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).