(a) To find: The force acting on the object at time t . Solution: F t = 2 m a i + 6 m b t j Explanation: 1) Concept: The force acting on object of mass m at time t is F t = m a s s × a c c e l e r a t i o n = m . a t 2) Given: r t = a t 2 i + b t 3 j , 0 ≤ t ≤ 1 . 3) Calculations: The position of an object at time t is r t = a t 2 i + b t 3 j Velocity is the rate of change of position with time. Therefore, v t = d r d t = d d t a t 2 i + b t 3 j = 2 a t i + 3 b t 2 j Acceleration is the rate of change of velocity with time. a t = d v d t = d d t 2 a t i + 3 b t 2 j = 2 a i + 6 b t j By using the concept, The force acting on object of mass m at time t is F t = m a s s × a c c e l e r a t i o n = m . a t F t = m 2 a i + 6 b t j = 2 m a i + 6 m b t j Conclusion: The force acting on the object at time t is F t = 2 m a i + 6 m b t j . (b) To find: The work done by the force during the time interval 0 ≤ t ≤ 1 . Solution: W = 2 m a 2 + 9 2 m b 2 Explanation: 1) Concept: The work done by the force field is W = ∫ c □ F · d r = ∫ a b F r t r ' t d t 2) Given: r t = a t 2 i + b t 3 j , 0 ≤ t ≤ 1 . 3) Calculation: From part (a), F r t = 2 m a i + 6 m b t j From given, r t = a t 2 i + b t 3 j Therefore, r ' t = 2 a t i + 3 b t 2 j . By using concept, the work done by the force during the time interval 0 ≤ t ≤ 1 is W = ∫ c □ F · d r = ∫ 0 1 F r t r ' t d t = ∫ 0 1 2 m a i + 6 m b t j · 2 a t i + 3 b t 2 j d t By using dot product, = ∫ 0 1 4 m a 2 t + 18 m b 2 t 3 d t By using power rule of integration, = 4 m a 2 t 2 2 + 18 m b 2 t 4 4 0 1 Simplify, = 2 m a 2 t 2 + 9 2 m b 2 t 4 0 1 Applying limits, = 2 m a 2 ( 1 ) 2 + 9 2 m b 2 1 4 - 0 = 2 m a 2 + 9 2 m b 2 Conclusion: The work done by the force during the time interval 0 ≤ t ≤ 1 is W = 2 m a 2 + 9 2 m b 2
(a) To find: The force acting on the object at time t . Solution: F t = 2 m a i + 6 m b t j Explanation: 1) Concept: The force acting on object of mass m at time t is F t = m a s s × a c c e l e r a t i o n = m . a t 2) Given: r t = a t 2 i + b t 3 j , 0 ≤ t ≤ 1 . 3) Calculations: The position of an object at time t is r t = a t 2 i + b t 3 j Velocity is the rate of change of position with time. Therefore, v t = d r d t = d d t a t 2 i + b t 3 j = 2 a t i + 3 b t 2 j Acceleration is the rate of change of velocity with time. a t = d v d t = d d t 2 a t i + 3 b t 2 j = 2 a i + 6 b t j By using the concept, The force acting on object of mass m at time t is F t = m a s s × a c c e l e r a t i o n = m . a t F t = m 2 a i + 6 b t j = 2 m a i + 6 m b t j Conclusion: The force acting on the object at time t is F t = 2 m a i + 6 m b t j . (b) To find: The work done by the force during the time interval 0 ≤ t ≤ 1 . Solution: W = 2 m a 2 + 9 2 m b 2 Explanation: 1) Concept: The work done by the force field is W = ∫ c □ F · d r = ∫ a b F r t r ' t d t 2) Given: r t = a t 2 i + b t 3 j , 0 ≤ t ≤ 1 . 3) Calculation: From part (a), F r t = 2 m a i + 6 m b t j From given, r t = a t 2 i + b t 3 j Therefore, r ' t = 2 a t i + 3 b t 2 j . By using concept, the work done by the force during the time interval 0 ≤ t ≤ 1 is W = ∫ c □ F · d r = ∫ 0 1 F r t r ' t d t = ∫ 0 1 2 m a i + 6 m b t j · 2 a t i + 3 b t 2 j d t By using dot product, = ∫ 0 1 4 m a 2 t + 18 m b 2 t 3 d t By using power rule of integration, = 4 m a 2 t 2 2 + 18 m b 2 t 4 4 0 1 Simplify, = 2 m a 2 t 2 + 9 2 m b 2 t 4 0 1 Applying limits, = 2 m a 2 ( 1 ) 2 + 9 2 m b 2 1 4 - 0 = 2 m a 2 + 9 2 m b 2 Conclusion: The work done by the force during the time interval 0 ≤ t ≤ 1 is W = 2 m a 2 + 9 2 m b 2
Solution Summary: The author explains that the force acting on the object at time t is F(t)=masstimes acceleration=m.
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