Two horizontal forces act on a 2.4 kg chopping block that can slide over a frictionless kitchen counter, which lies in an xy plane. One force is F1 = (3.4 N) i + (3.7 N) j. Find the acceleration of the chopping block in unit-vector notation for each of the following second forces:
F2 = (-3.4 N) i + (-3.7 N) j
F2 = (-3.4 N) i + (3.7 N) j
F2 = (3.4 N) i + (-3.7 N) j
In vector notation;
Force= mass*acceleration
acceleration= Force/mass
add the vector forces, divide bymass
example, the last:
Ftotal=F1 + F2
= 3.4 i + 3.7 j + 3.4i - 3.7 j
= 6.8 i
acceleration= 6.8/2.4 i
a = (F1 + F2) / m
where a, F1, & F2 are in vector notation
To find the acceleration of the chopping block in unit-vector notation, we need to calculate the total force acting on the block and then divide it by the mass of the block.
Let's solve each of the given scenarios step by step:
1. F2 = (-3.4 N)i + (-3.7 N)j:
To find the total force (Ftotal), we need to sum up the individual forces (F1 and F2):
Ftotal = F1 + F2
= (3.4 N)i + (3.7 N)j + (-3.4 N)i + (-3.7 N)j
= 0i + 0j
= 0 N
Since the total force is zero, the acceleration would be zero as well. Thus, the acceleration in unit-vector notation for this scenario would be 0i.
2. F2 = (-3.4 N)i + (3.7 N)j:
Again, let's calculate the total force:
Ftotal = F1 + F2
= (3.4 N)i + (3.7 N)j + (-3.4 N)i + (3.7 N)j
= 0i + 7.4j
= 7.4 Nj
The total force is 7.4 N in the positive y-direction. Now, let's find the acceleration:
acceleration = Ftotal / mass
= (7.4 Nj) / 2.4 kg
= (0i + 7.4j) / 2.4
= (0/2.4)i + (7.4/2.4)j
= 0i + 3.08j
Therefore, the acceleration in unit-vector notation for this scenario is 0i + 3.08j.
3. F2 = (3.4 N)i + (-3.7 N)j:
Let's calculate the total force:
Ftotal = F1 + F2
= (3.4 N)i + (3.7 N)j + (3.4 N)i + (-3.7 N)j
= 6.8i + 0j
= 6.8 N i
The total force is 6.8 N in the positive x-direction. Now, let's find the acceleration:
acceleration = Ftotal / mass
= (6.8 Ni) / 2.4 kg
= (6.8/2.4)i + (0/2.4)j
= 2.83i + 0j
Hence, the acceleration in unit-vector notation for this scenario is 2.83i + 0j.