I think that in order for work to be done, the object must move in the direction of the force and move over a distance.
Question:
A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
(a)60 (b)90 (c)120
Answer:
(a)5.42 N (b)6.26 N (c)5.42 N
Explanation:
From the question
Length of wire (L) = 2.80 m
Current in wire (I) = 5.20 A
Magnetic field (B) = 0.430 T
Angle are different in each part.
The magnetic force is given by

So from data

Now sub parts
(a)

(b)

(c)

Answer:
W ≅ 292.97 J
Explanation:
1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)
Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;
W = (Fcosθ)d
Given that:
Tension of the force = 62 N
angle of incline θ = 34°
distance d =5.7 m.
Then;
W = 62 × cos(34) × 5.7
W = 353.4 cos(34)
W = 353.4 × 0.8290
W = 292.9686 J
W ≅ 292.97 J
Hence, the work done by tension before the block goes up the incline = 292.97 J
Answer:
Option D
670 Kg.m/s
Explanation:
Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)
Final momentum is also mv but v being westward direction, we take it negative
Final momentum=82*-2.5= -205 Kg.m/s
Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s
Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s