×
×

# Consider the blocks in Exercise 6.7 as they move 75.0 cm. ISBN: 9780321675460 31

## Solution for problem 66P Chapter 6

University Physics | 13th Edition

• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants University Physics | 13th Edition

4 5 1 284 Reviews
28
0
Problem 66P

Consider the blocks in Exercise 6.7 as they move 75.0 cm. Find the total work done on each one (a) if there is no friction between the table and the 20.0-N block, and (b) if µs = 0.500 and µk = 0.325 between the table and the 20.0-N block. 6.7 . Two blocks are connected by a very light string passing over a massless and frictionless pulley (?Fig. E6.7?). Traveling at constant speed, the 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm down-ward. How much work is done (a) on the 12.0-N block by (i) gravity and (ii) the tension in the string? (b) How much work is done on the 20.0-N block by (i) gravity, (ii) the tension in the string, (iii) friction, and (iv) the normal force? (c) Find the total work done on each block.

Step-by-Step Solution:

Solution 66P a) Without friction: Step 1: The work done by gravity on block 1 is, W = F gravitydisplacement = force × displacement × cos Here gravity is acting vertically downwards and the block is moving horizontally. The angle between them is 90 degree, so the cosine of the angle is zero. Hence the work done is zero. Step 2: Work done by the frictional force is also zero as there is no friction between the block and the plane. The total work done on the 1st block is zero. Step 3: The block of 12 N is moving vertically downwards, where the gravitational force is also in the same direction. So, the work done by gravity on the the 2nd block is, W = 20N × 0.75m × cos 0 = 15 jules. Again as there is no friction, the work done by friction is zero. The total work done is 15 jules. b) when there is friction: Step 4: The kinetic and static friction of the 20 N block are respectively, k 0.325 and = s.500. The frictional force is, ( + ) × normal force = (0.325 + 0.500) × 20 = 16.5 N k s Work done on block 1 by frictional force is, W = 16.5 × 0.75 = 12.375 joules.

Step 5 of 7

Step 6 of 7

##### ISBN: 9780321675460

Unlock Textbook Solution