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The free-body diagrams for block B and for the knot just above block A are shown below. $$\vec{T_1}$$ is the tension force of the rope pulling on block B or pulling on the knot (as the case maybe), $$\vec{T_2}$$ is the tension force exerted by the second rope (at angle $$\theta = 30^\circ$$) on the knot, $$\vec f$$ is the force of static friction exerted by the horizontal surface on block B ...
Assertion :Block a is moving on the horizontal surface towards right under action of force F. all surfaces are smooth. At the instant shown, the force exerted by block A on block B is equal to net force on block B. Reason: From Newton's third law, the force
Block A weight 4N and block weight 8N. The coefficient of kinetic friction is 0.25 for all surfaces. Find the force F to slide B at a constant speed when (a0 A rests on B and moves with it, (b) A is held at rest, and (c) A and B are connected by a light cord passing
Click here:point_up_2:to get an answer to your question :writing_hand:block b moves without rotation vertically downwards with constant velocity of ms then what is 67.Block ‘B’ moves without rotation vertically downwards with constant velocity of 1m/s then what is the relative velocity of C with respect to A :
Speed amplifier. In Fig.$$ 9-75 $$ , block $$ 1 $$ of mass $$ m_1 $$ slides along an x axis on a frictionless floor at speed $$ \nu_{1i} = 4.00\, m/s. $$ Then it undergoes a one-dimensional elastic collision with stationary block $$ 2 $$ of mass $$ m_2 = 0.500m_1.$$ Next, block $$ 2 $$ undergoes a one-dimensional elastic collision with stationary block $$ 3 $$ of mass $$ m_3 = 0.500m_2 $$ .
In Fig. $$8-60,$$ the pulley has negligible mass, and both it and the inclined plane are frictionless. Block A has a mass of $$1.0\, kg,$$ block B has a mass of $$2.0\, kg,$$ and angle $$\theta$$ is $$30^\circ.$$ If the blocks are released from rest with the connecting cord taut, what is their total kinetic energy when block B has fallen $$25\, cm$$?
A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a horizontal floor. The coefficient of friction between A and B is 0.2, while that between B and floor is 0.5. When a horizontal force of 45 N is applied on the block B, the force of friction
Block B has mass m and is released from rest when it is on top of wedge A, which has a mass 3m. Determine the tension in cord CD needed to hold the wedge from moving while B is sliding down A. Neglect friction.
Block A of mass 2 k g is placed over a block B of mass 8 k g. The combination is placed on a rough horizontal surface. If g = 10 m s − 2, coefficient of friction between B and floor = 0.5, coefficient of friction between A and B = 0.4 and a horizontal force of 10 N 8
Click here:point_up_2:to get an answer to your question :writing_hand:23block b moves without rotation vertically downwards with constant velocity of 1ms then what isthe A small ball falling vertically downward with constant velocity 4 m/s strikes elastically a massive wedge moving with velocity 4 m/s horizontally as shown.
