Answer:
Step-by-step explanation:
(a) The kinetic energy of the object at time t1 is given by:
K1 = (1/2)mv1^2
where m = 8.51 kg and v1 = (5.21î − 2.81ĵ) m/s. Substituting these values, we get:
K1 = (1/2)(8.51 kg)(5.21î − 2.81ĵ)^2 = 149.12 J
Therefore, the kinetic energy of the object at time t1 is 149.12 J.
(b) The displacement of the object during the time interval Δt = t2 − t1 is given by:
Δr = r2 − r1 = (8.00î + 4.00ĵ) m - 0î - 0ĵ = 8.00î + 4.00ĵ
The average velocity of the object during this time interval is:
vavg = Δr/Δt = (8.00î + 4.00ĵ)/(2.00 s - 0 s) = 4.00î + 2.00ĵ m/s
Using the kinematic equation:
Δv = aΔt
where Δv = v2 - v1 is the change in velocity during the time interval, we can solve for the acceleration:
a = Δv/Δt = (v2 - v1)/Δt = [(8.00î + 4.00ĵ) m/s - (5.21î − 2.81ĵ) m/s]/(2.00 s - 0 s) = 1.395î + 3.415ĵ m/s^2
The force acting on the object during the time interval Δt = t2 − t1 is given by Newton's second law:
F = ma = (8.51 kg)(1.395î + 3.415ĵ) m/s^2 = 11.83î + 29.05ĵ N
Therefore, the force acting on the object during the time interval Δt is F = 11.83î + 29.05ĵ N.
(c) The work done on the object by the force during the time interval Δt = t2 − t1 is given by:
W = F · Δr
where · denotes the dot product. Substituting the values of F and Δr, we get:
W = (11.83î + 29.05ĵ) N · (8.00î + 4.00ĵ) m = 95.52 J
Therefore, the work done on the object by the force during the time interval Δt is -95.52 J.
Note: The negative sign indicates that the work done by the force is negative, i.e., the force acts in the opposite direction to the displacement of the object.
(d) The kinetic energy of the object at time t2 is given by:
K2 = (1/2)mv2^2
where m = 8.51 kg and v2 is the velocity of the object at time t2. To find v2, we can use the kinematic equation:
v2 = v1 + aΔt
where a is the acceleration of the object during the time interval and Δt = t2 − t1. Substituting the values, we get:
v2 = (5.21î − 2.81ĵ) m/s + (1.395î + 3.415ĵ) m/s^2(2.00 s)