We show below a free-body diagram of a 1.00-kg block pulled across a surface by

Question

We show below a free-body diagram of a 1.00-kg block pulled across a surface by a string.  The string's tension is T = 6 N.  The string makes an angle of 37 degrees with the horizontal.  The coefficient of kinetic friction is uk = 0.30.

1. determine the x-component of the force of friction?

2. find the magnitude of the total force on the block?

We show below a free-body diagram of a 1.00-kg block pulled across a surface by a string. The string's tension is T = 6 N. The string makes an angle of 37 degrees with the horizontal. The coefficient of kinetic friction is uk = 0.30. determine the x-component of the force of friction? find the magnitude of the total force on the block?

Explanation / Answer

(a) we have Frictional force = uk x Normal Reaction

here, Normal Reaction = T sin 37 - mg = - 6.2 N i.e downwards

SO, frictional force = 0.3 x 6.2 = 1.86 N

(b) Total Force on the block

Vertical Force = mg - T sin 37 = 6.3 N

Horizontal Force = T cos 37 - fk = 4.788 - 1.86 = 2.928

Total Force = sqrt(6.3^2 + 2.928^2) = 6.947 N

Answer is 6.947 N

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