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SOLUTION OF ASSIGNMENT NO. ( 5 )

# SOLUTION OF ASSIGNMENT NO. ( 5 ) - 1 1 SOLUTION OF...

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Unformatted text preview: 1 1 SOLUTION OF ASSIGNMENT NO. ( 5 ) – KINETICS OF A PARTICLE FORCE AND ACCELERATION NEWTON` S SECOND LAW OF MOTION 1. The 50 – kg crate shown in Fig. 1, rests on a horizontal plane for which the coefficient of kinetic friction μ = 0.3. If the crate is subjected to a P = 400 N towing force as shown, determine the velocity of the crate in 3 seconds starting from rest. Solution Kinetics 2 o o o Y o K o K X s / m 13 . 5 50 30 cos 400 ) 300 ( 3 . a : ) 1 ( in ) 2 ( from ng Substituti ) 2 ( ...... N 300 ) 5 . ( 400 500 30 sin P mg N 30 sin P mg N ve F ) 1 ( . .......... m 30 cos P N a ma 30 cos P N ve ma F = + − = = − = − = ⇒ = + − ⊕ ↑ = + − = ⇒ = + − ⊕ → = ∑ ∑ μ μ F r =µ K N External Forces ° 30 P N m g Pcos30 o Psin30 o Effective Forces m a 2 2 Kinematics Since a = constant, we can use : v=v o + a t , where v o = 0 since the crate started from rest. Therefore the velocity after 3 seconds is: v = (5.13) × (3) = 15.39 m/s Or we can use the general form: s / m 39 . 15 v dv dt 13 . 5 dv adt dt dv a 3 v t v = ⇒ = ⇒ = ⇒ = ∫ ∫ ∫ ∫ 2. A mine car has a mass of 400 kg is subjected to a tension force, exerted by a motor M , which is varying with the time P = 3200 t 2 N , where t is in seconds. If the car was initially moving with an upward velocity of v o = 2 m / s , determine its velocity and position after two seconds. (g = 10 m / s 2 , tanα = 8/15 ) . Solution N 41 . 3529 17 15 4000 cos g m N cos g m N F ) 1 ...( .......... 17 80 t 8 a 400 17 8 4000 t 3200 a m sin g m P a a m sin g m P a m F Y 2 2 X =       × = = ⇒ = − =       − = ⇒       × − = ⇒ − = ⇒ = − ⇒ = ∑ ∑ α α α α Kinematics α m a Effective Forces m g External Forces α P M α N mgcos α m g sin α M 3 3 As shown in equation (1) the acceleration is not constant, therefore we have to integrate to get both the velocity and position of the mine car: ) 3 .( .......... .......... t 2 t 17 40 t 3 2 x t d 2 t 17 80 t 3 8 x d x d t d v dt dx v ) 2 ( .......... 2 t 17 80 t 3 8 v t 17 80 t 3 8 2 v t d 17 80 t 8 dv dv dt a dt dv a 2 4 t 3 x 3 3 t 2 v 2 + − = ⇒  ...
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SOLUTION OF ASSIGNMENT NO. ( 5 ) - 1 1 SOLUTION OF...

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